Question 13 Marks
Find the values of k for which the given quadratic equation has real and distinct root:
$kx^2 + 2x + 1 = 0$
Answer$k x^2+2 x+1=0$
Here, $a=k, b=2, c=1$
$\therefore \text { Discriminant }(D)=b^2-4 a c$
$=(2)^2-4 \times k \times 1$
$=4-4 k$
$\because$ Roots are real and distinct,
$\therefore D>0$
$\Rightarrow 4-4 k>0$
$\Rightarrow 1-k>0$
$\Rightarrow 1>k$
$\Rightarrow k<1$
$\therefore k<1$
View full question & answer→Question 23 Marks
Find the value of k for which the following equation have real root:
$4x^2 + kx + 3 = 0$
AnswerThe given quadration is $4x^2 + kx + 3 = 0$, and roots are real and equal.Then find the value of p.
Here, $4x^2 + kx + 3 = 0$
So, $a = 4, b = p$ and $c = 3$
As know that $D = b^2 - 4ac$
Putting the value of $a = 4, b = p$ and $c = 3$
$D = (p)^2 - 4(4)(3)$
$D = p^2 - 48$
The given equation will have real and equal roots, if D = 0
So, $p2 - 48 = 0$
Now factorizing the above equation,
$p2 - 48 = 0$
$\Rightarrow\text{p}^2-(4\sqrt{3})^2=0$
$\Rightarrow(\text{p}-4\sqrt{3})(\text{p}+4\sqrt{3})=0$
$\Rightarrow\text{p}-4\sqrt{3}=0$ or $\text{p}+4\sqrt{3}=0$
$\Rightarrow\text{p}=4\sqrt{3}$ or $\text{p}=-4\sqrt{3}$
Therefore, the value of $\text{p}=\pm4\sqrt{3}$
View full question & answer→Question 33 Marks
Find the least positive value of $k$ for which the equation $x^2 + kx + 4 = 0$ has real roots.
AnswerThe given equation is $x^2 + kx + 4 = 0$
Given that the equation has real roots
i.e., $\text{D}=\text{b}^2-4\text{ac}\geq0$
$\Rightarrow\text{k}^2-4\times1\times4\geq0$
$\Rightarrow\text{k}^2-16\geq0$
$\Rightarrow\text{k}\geq4$ or $\text{k}\leq-4$
The least positive value of $k = 4$, for the equation to have real roots.
View full question & answer→Question 43 Marks
Find the value of k for which the root are real and equal in the following equations:
$4x^2 + kx + 9 = 0$
AnswerThe given equation is $4 x^2+k x+9=0$
This equation is in the form of $a x^2+b x+c=0$
Here, $a=4, b=k$ and $c =9$
Given that, the equation has real and equal roots
$\text { i.e., } D=b^2-4 a c=0$
$\Rightarrow k^2-4 \times 4 \times 9=0$
$\Rightarrow k^2-16 \times 9$
$\Rightarrow k=\sqrt{16 \times 9}$
$\Rightarrow k=4 \times 3$
$\Rightarrow k=12$
$\therefore$ The value of $k =12$
View full question & answer→Question 53 Marks
Find the value of k for which the following equation have real root:
$kx(x - 3) + 9 = 0$
Answer$k x(x-3)+9=0$
$k x^2-3 k x+9=0$
Here, $a=k, b=-3 k, c=9$
$\therefore D=b^2-4 a c$
$\Rightarrow D=(-3 k)^2-4(k) 9$
$\Rightarrow D=9 k^2-36 k$
For roots to be real
$\Rightarrow D=0$
$\Rightarrow 9 k^2-36 k=0$
$\Rightarrow 9 k(k-4)=0$
$\Rightarrow k-4=0$
$\Rightarrow k=4$
$\therefore k=4$
View full question & answer→Question 63 Marks
In the following, determine whether the given values are solution of the given equation or not:
$x^2 + x + 1 = 0, x = 0, x = 1$
AnswerWe have been given that,
$x^2 + x + 1 = 0, x = 0, x = 1$
Now, if $x = 0$ is a solution of the equation then it should satisfy the equation.
So, substituting $x = 0$ in the equation we get
$x^2 + x + 1$
$= (0)^2 + (0) + 1$
$= 1$
Hence $x = 0$ is not a solution of the given quadratic equation.
Also, if $x = 1$ is a solution of the equation it should satisfy the equation So, substituting $x = 1$ in the equation, we get
$x^2 + x + 1$
$= (1)^2 + (1) + 1$
$= 3$
Hence $x = 1$ is not a soluion of the quadratic equation.
Therefore, from the above results we find out that both $x = 0$ and $x = 1$ are not a solution of the given quadratic equation.
View full question & answer→Question 73 Marks
Solve the following quadratic equations by factorization:$(a + b)^2 x^2 - 4abx - (a - b)^2 = 0$
AnswerWe have been given
$(a + b)^2 x^2 - 4abx - (a - b)^2 = 0$
$(a + b)^2 x^2 - (a + b)^2 x + - (a - b)^2 = 0$
$(a + b)^2 x(x - 1) + (a - b)^2 (x - 1) = 0$
$((a + b)^2 x + (a + b)^2) (x - 1) = 0$
Therefore,
$(a + b)^2 x + (a - b)^2 = 0$
$(a + b)^2 x = -(a - b)^2$
$\text{x}=\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\Big)^2$
or, $x - 1 = 0$
$x = 1$
Hence, $\text{x}=\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\Big)^2$ or $x = 1$
View full question & answer→Question 83 Marks
If an integer is added to its square, the sum is $90$. Find the integer with the help of quadratic equation.
AnswerLet the integer be $'x'$
Given that if an integer is added to its square, the sum is 90
$\Rightarrow x + x^2 = 90$
$\Rightarrow x + x^2 - 90 = 0$
$\Rightarrow x^2 + 10x - 9x - 90 = 0$
$\Rightarrow x(x + 10) - 9(x + 10) = 0$
$\Rightarrow x = -10$ or $x = 9$
$\therefore$ The value of an integer are $-10 $or $9$
View full question & answer→Question 93 Marks
An express train takes 1 hour less than a passenger train to travel 132km between Mysore and Banglore. If the average speed of the express train is 11km/ hr more than that of the passenger train, from the quadratic equation to find the average speed of express train.
AnswerNow let us assume taht the speed of the speed of the express train be 'x' km/ hr. Therefore according to the question spreed of the passenger train will be 'x - 11'km/ hr. Now we know that the total distance travelled by both rthe trains was 132km.
We also know that so, the time by express train would be $\Big(\frac{132}{\text{x}}\Big)$ hr and the time taken by the passenger train would be $\Big(\frac{132}{\text{x}-11}\Big)$ he. Now, we also know that the express train took 1 hr less than the passenger.
Therefore, we have
$\Big(\frac{132}{\text{x}}\Big)=\Big(\frac{132}{\text{x}-11}\Big)-1$
$\Big(\frac{132}{\text{x}-11}\Big)-\Big(\frac{132}{\text{x}}\Big)=1$
$\Big(\frac{132\text{x}-132(\text{x}-11)}{\text{x}^2-11\text{x}}\Big)=1$
$\text{x}^2-11\text{x}-1452=0$
Therefore, this is the required equation.
View full question & answer→Question 103 Marks
Determine. if 3 is a root of the equation given below:
$\sqrt{\text{x}^2-4\text{x}+3}+\sqrt{\text{x}^2-9}=\sqrt{4\text{x}^2-14\text{x}+16}$
Answer$\sqrt{\text{x}^2-4\text{x}+3}+\sqrt{\text{x}^2-9}=\sqrt{4\text{x}^2-14\text{x}+16}$
If x = 3, then
L.H.S. $\sqrt{\text{x}^2-4\text{x}+3}+\sqrt{\text{x}^2-9}$
$=\sqrt{(3)^2-4\times3+3}+\sqrt{(3)^2-9}$
$=\sqrt{9-12+3}+\sqrt{9-9}$
$=\sqrt{0}+\sqrt{0}$
$=0$
R.H.S. $=\sqrt{4\text{x}^2-14\text{x}+16}$
$=\sqrt{4(3)^2-14\times3+16}$
$=\sqrt{4\times9-42+16}$
$=\sqrt{36+16-42}$
$=\sqrt{52-42}$
$=\sqrt{10}$
$\because$ L.H.S. $\neq$ R.H.S.
View full question & answer→Question 113 Marks
Write the value of $\lambda,$ for which $\text{x}^2+4\text{x}+\lambda$ is a perfect square.
AnswerIn $\text{x}^2+4\text{x}+\lambda$
$\text{a}=1,\text{b}=4,\text{c}=\lambda$
$\text{x}^2+4\text{x}+\lambda$ will be a perfect square if $\text{x}^2+4\text{x}+\lambda=0$ has equal roots
$\Rightarrow\text{D}=\text{b}^2-4\text{ac}$
$\Rightarrow\text{D}=(4)^2-4\times1\times\lambda$
$\Rightarrow\text{D}=0$
$\Rightarrow16-4\lambda=0$
$\Rightarrow16=4\lambda$
$\Rightarrow\lambda=4$
Hence $\lambda=4$
View full question & answer→Question 123 Marks
Solve the following quadratic by factorization:$a(x^2 + 1) - x(a^2 + 1) = 0$
AnswerWe have
$a\left(x^2+1\right)-x\left(a^2+1\right)=0$
$\Rightarrow a\left(x^2-1\right)-a^2 x-x+a=0$
${\left[\because a \times a=a^2 \Rightarrow a^2=-a^2 \times-1-\left(a^2+1\right)=a^2-1\right]}$
$\Rightarrow a \times(x-a)-1(x-a)=0$
$\Rightarrow(x-a)(a x-1)=0$
$\Rightarrow x-a=0 \text { or } a x-1=0$
$\Rightarrow x=a \text { or } x=\frac{1}{a}$
$\therefore x=a$ and $x=\frac{1}{a}$ are the two roots of the given equation.
View full question & answer→Question 133 Marks
The difference of two natural numbers is $3$ and the difference of their reciprocals is $\frac{3}{28}.$ Find the numbers.
AnswerLet the larger tatural number be $x$
and the smaller natural number be $y$
$\Rightarrow x - y = 3 .....(i)$
$\Rightarrow x = 3 + y ......(ii)$
$\Rightarrow\frac{1}{\text{y}}-\frac{1}{\text{x}}=\frac{3}{28}$ $\Big[$ If x < y then $\frac{1}{\text{x}}<\frac{1}{\text{y}}\Big]$
$\Rightarrow\frac{\text{x}-\text{y}}{\text{xy}}=\frac{3}{28}$
$\Rightarrow\frac{3}{\text{xy}}=\frac{3}{28}$ [From (i)]
$\Rightarrow\frac{1}{\text{xy}}=\frac{1}{28}$
$\Rightarrow xy = 28$
$\Rightarrow (3 + y)y = 28 [From (ii)]$
$\Rightarrow 3y + y^2 - 28 = 0$
$\Rightarrow y^2 + 3y - 28 = 0$
$\Rightarrow y^2 + 7y - 4y - 28 = 0$
$\Rightarrow y(y + 7) - 4(y + 7) = 0$
$\Rightarrow (y - 4)(y + 7) = 0$
$\Rightarrow y = 4$ or $y = -7$ (rejected being natural no.)
When $y = 4, x = 3 + 4 = 7$ [From (ii)]
Number are $7, 4$
View full question & answer→Question 143 Marks
Find the consecutive even integers whose squares have the sum $340$.
AnswerLet the consecutive even integers be $2 x$ and $2 x+2$
Then according to the given hypothesis,
$(2 x)^2+(2 x+2)^2=340$
$4 x^2+4 x^2+8 x+4-340=0$
$\Rightarrow 8 x^2+8 x-336=0$
$\Rightarrow x^2+x-42=0$
$\Rightarrow x^2+7 x-6 x-42=0$
$\Rightarrow x(x+7)-6(x+7)=0$
$\Rightarrow(x+7)(x-6)=0$
$\Rightarrow x=-7 \text { or } x=6$
Considering, the positive integers of $x=6$
$\Rightarrow 2 x=12 \text { and } 2 x+2=14$
$\therefore$ The two consecutive even integers are $12$ and $14$
View full question & answer→Question 153 Marks
Determine the set of values of k for which the given following quadratic has real root:
$x^2 - kx + 9 = 0$
Answer$x^2 - kx + 9 = 0$
Here, $a = 1, b = -k, c = 9$
$\therefore$ Discriminant $(D) = b^2 - 4ac$
$= (-k)^2 - 4 \times 1 \times 9$
$= k^2 - 36$
$\because$ Roots are real
$\therefore\text{D}\geq0$
$\Rightarrow\text{k}^2-36\geq0$
$\Rightarrow\text{k}^2\geq36$
$\Rightarrow\text{k}^2\geq(\pm6)^2$
$\therefore\text{k}\geq6$ or $\text{k}\leq-6$
View full question & answer→Question 163 Marks
Determine the nature of the root of following quadratic equation:
$(x - 2a)(x - 2b) = 4ab$
Answer$(x - 2a)(x - 2b) = 4ab$
$\Rightarrow x^2 - 2bx - 2ax + 4ab - 4ab = 0$
$\Rightarrow x^2 - 2(a + b)x = 0$
Here $a = 1, b = -2(a + b), c = 0$
$\therefore$ Discriminant $(D) = b^2 - 4ac$
$= {-2(a + b)}^2 - 4 \times 1 \times 0$
$= {-2(a + b)}^2$
$\because$ $D > 0$
$\therefore$ Roots are real and distinct.
View full question & answer→Question 173 Marks
The sum of two numbers is $9$. The sum of their reciprocals is $\frac{1}{2}.$ Find the numbers.
AnswerGiven that the sum of two number is $9$ let the two numbers be $x$ and $9 - x$
By the given hypothesis, we have
$\frac{1}{\text{x}}+\frac{1}{9-\text{x}}=\frac{1}{2}$
$\Rightarrow\frac{9-\text{x}+\text{x}}{\text{x}(9-\text{x})}=\frac{1}{2}$
$\Rightarrow 18 = 9x - x^2$
$\Rightarrow x^2 - 9x + 18 = 0$
$\Rightarrow x^2 - 6x - 3x + 18 = 0$
$\Rightarrow x(x - 6) - 3(x - 6) = 0$
$\Rightarrow (x - 6)(x - 3) = 0$
$\Rightarrow x = 6$ or $x = 3$
$\therefore$ The two numbers are $3$ and $6$
View full question & answer→Question 183 Marks
Determine the set of values of k for which the given following quadratic has real root:
$2x^2 + kx - 4 = 0$
Answer$2x^2 + kx - 4 = 0$
Here, $a = 2, b = k, c = -4$
$\therefore$ Discriminant $(D) = b^2 - 4ac$
$= k^2 - 4 \times 2 \times (-4)$
$= k^2 + 32$
$\because$ The roots are real
$\therefore\text{D}\geq0$
$\Rightarrow\text{k}^2+32\geq0$
$\because\text{k}^2+32\geq0$ for all value of $\text{k }\in\text{ R}$
View full question & answer→Question 193 Marks
The sum of the squares of two numbers is $233$ and one of the number is $3$ less than twice the other number. Find the numbers.
AnswerLet the number be $x$
Then the other number $= 2x - 3$
According to the given hypothesis,
$\Rightarrow x^2 + (2x - 3)^2 = 233$
$\Rightarrow x^2 + 4x^2 + 9 - 12x = 233$
$\Rightarrow 5x^2 - 12x - 224 = 0 .....(i)$
The value of $'x'$ can obtained by the formula $\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
Here, $a = 5, b = -12$ and $c = -224$ from (i)
$\Rightarrow\text{x}=\frac{-(-12)+\sqrt{144+20\times224}}{10}=8$
$\Rightarrow\text{x}=\frac{-(-12)-\sqrt{144+20\times224}}{10}=\frac{-28}{5}$
Considering the value of $x = 8$
$\Rightarrow 2x - 3$
$\Rightarrow 2 \times 8 - 3$
$\Rightarrow 16 - 3$
$\Rightarrow 13$
$\therefore$ The two number are $8$ and $13$.
View full question & answer→Question 203 Marks
Find the values of k for which the given quadratic equation has real and distinct root:
$kx^2 + 6x + 1 = 0$
Answer$kx^2 + 6x + 1 = 0$
Here, $a = k, b = 6, c = 1$
$\therefore$ Discriminant $(D) = b^2 - 4ac$
$= (6)^2 - 4 \times k \times 1$
$= 36 - 4k$
$\because$ Roots are real and distinct,
$\therefore$$ D > 0$
$\Rightarrow 36 - 4k > 0$
$\Rightarrow 9 - k > 0$
$\Rightarrow 9 > k$
$\Rightarrow k < 9$
$\therefore$ $k < 9$
View full question & answer→Question 213 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$3x^2 - 5x + 2 = 0$
Answer$3x^2 - 5x + 2 = 0$
Here, $a = 3, b = -5, c = 2$
$D = b^2 - 4ac$
$= (-5)^2 - 4 \times 3 \times 2$
$= 25 - 24$
$= 1$
$\therefore$ $D > 0$
$\therefore$ Roots are real
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$=\frac{-(-5)\pm\sqrt{1}}{2\times3}=\frac{5\pm1}{6}$
$\therefore\text{x}=\frac{5+1}{6}=\frac{6}{6}=1$
or $\text{x}=\frac{5-1}{6}=\frac{4}{6}=\frac{2}{3}$
$\text{x}=1,\frac{2}{3}$
View full question & answer→Question 223 Marks
Find two natural numbers which differ by $3$ and whose squares have the sum $117$.
AnswerLet first number $= x$
Then second number $=x-3$
According to the condition,
$x^2+(x-3)^2=117$
$\Rightarrow x^2+x^2-6 x+9=117$
$\Rightarrow 2 x^2-6 x+9-117=0$
$\Rightarrow 2 x^2-6 x-108=0$
$\left.\Rightarrow x^2-3 x-54=0 \text { (Dividing by } 2\right)$
$\Rightarrow x^2-9 x+6 x-54=0$
$\Rightarrow x(x-9)+6(x-9)=0$
$\Rightarrow(x-9)(x+6)=0$
Either $x-9=0$, then $x=9$
Or $x+6=0$, then $x=-6$ which is not a natural number,
First natural number $=9$
and second number $=9-3=6$
View full question & answer→Question 233 Marks
Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is $20$ years ago, the product of their ages in years was $48$.
AnswerSum of ages of two friends = $20$ years
Let present age of one friend = $x$ years
Age of second friend = $(20 - x)$ years
$4$ years ago,
Age of first friend = $x - 4$
and age of second friend $= 20 - x - 4 = 16 - x$
According to the condition,
$(x - 4)(16 - x) = 48$
$\Rightarrow 16x - x^2 - 64 + 4x = 48$
$\Rightarrow -x^2 + 20x - 64 - 48 = 0$
$\Rightarrow -x^2 + 20x - 112 = 0$
$\Rightarrow x^2 - 20x + 112 = 0$
Here $a = 1, b = -20, c = 112$
Discriminan $(D) = b^2 - 4ac$
$= (-20)^2 - 4 \times 1 \times 112$
$= 400 - 448 = -48$
$\therefore$ $D < 0$
Roots are not real
It is not possible.
View full question & answer→Question 243 Marks
If $1$ is a root of the quadratic equation $3x^2 + ax - 2 = 0$ and the quadratic equation $a(x^2 + 6x) - b = 0$ has equal roots, find the value of b.
Answer$1$ is one root of $3x^2 + ax - 2 = 0$
$\therefore 3(1)^2 + a \times 1 - 2 = 0$
$\Rightarrow 3 + a - 2 = 0$
$\Rightarrow a + 1 = 0$
$\Rightarrow a = -1$
Now in equation $a(x^2 + 6x) - b = 0$
$\Rightarrow -1(x^2 + 6x) - b = 0$
$\Rightarrow -x^2 - 6x - b = 0$
$\Rightarrow x^2 + 6x + b = 0$
Here $A = 1, B = 6, C = b$
$\therefore$ $D = B^2 - 4AC$
$\Rightarrow D = (6)^2 - 4 \times 1 \times k$
$\Rightarrow D = 36 - 4k$
$\because$ Roots are equal
$\Rightarrow D or B^2 - 4AC = 0$
$\Rightarrow 36 - 4k = 0$
$\Rightarrow 4k = -36$
$\Rightarrow\text{k}=\frac{-36}{4}=-9$
View full question & answer→Question 253 Marks
Solve the following quadratic equations by factorization:
$\frac{1}{\text{x}-1}-\frac{1}{\text{x}+5}=\frac{6}{7},$ $\text{x}\neq1,-5$
Answer$\frac{1}{\text{x}-1}-\frac{1}{\text{x}+5}=\frac{6}{7}$ $(\text{x}\neq1,-5)$
$\frac{\text{x}+5-\text{x}+1}{(\text{x}-1)(\text{x}+5)}=\frac{6}{7}$
$\Rightarrow\frac{6}{(\text{x}-1)(\text{x}+5)}=\frac{6}{7}$
$\Rightarrow\frac{1}{(\text{x}-1)(\text{x}+5)}=\frac{1}{7}$ (Dividing by 6)
$(x - 1)(x + 5) = 7 \Rightarrow x^2 + 5x - x - 5 = 7$
$\begin{Bmatrix}\because-12=6\times(-2)\\+4=6-2\end{Bmatrix}$
$\Rightarrow x^2 + 4x - 5 - 7 = 0$
$\Rightarrow x^2 + 4x - 12 = 0$
$\Rightarrow x^2 + 6x - 2x - 12 = 0$
$\Rightarrow x(x + 6) - 2(x + 6) = 0$
$\Rightarrow (x + 6)(x - 2) = 0$
Either $x + 6 = 0$, then $x = -6$
or $x - 2 = 0$, then $x = 2$
$\therefore$ $x = 2, -6$
View full question & answer→Question 263 Marks
Find the value of $k$ for which the following equations have real and equal roots:
$x^2+k(2 x+k-1)+2=0$
AnswerThe given equation is $x^2+k(2 x+k-1)+2=0$
$\Rightarrow x^2+2 k x+k(k-1)+2=0$
So, $a=1, b=2 k, c=k(k-1)+2$
We know $D=b^2-4 a c$
$\Rightarrow D=(2 k)^2-4 \times 1 \times[k(k-1)+2]$
$\Rightarrow D=4 k^2-4\left[k^2-k+2\right]$
$\Rightarrow D=4 k^2-4 k 2+4 k-8$
$\Rightarrow D=4 k-8$
$\Rightarrow D=4(k-2)$
For equal roots, $D =0$
Thus, $4(k-2)=0$
So, $k=2$
View full question & answer→Question 273 Marks
Find the value of k for which the root are real and equal in the following equations:
$\text{kx}^2-2\sqrt{5}\text{x}+4=0$
AnswerThe given equation $\text{kx}^2-2\sqrt{5}\text{x}+4=0$ is in the from of $ax^2 + bx + c = 0$ where
$\text{a}=\text{k},\text{b}=-2\sqrt{3}$ and $\text{c}=4$
Gievn that, the equation has and equal roots,
i.e., $\text{D}=\text{b}^2-4\text{ac}=0$
$\Rightarrow(-2\sqrt{5})^2-4\times\text{k}\times4=0$
$\Rightarrow20=16\text{k}$
$\Rightarrow\text{k}=\frac{20}{16}=\frac{5}{4}$
$\therefore\text{k}=\frac{5}{4}$
$\therefore$ The value of $\text{k}=\frac{5}{4}$
View full question & answer→Question 283 Marks
Find the value of k for which root are real and equal in the following equations:
$(k + 1)x^2 + 2(k + 3)x + (k + 8) = 0$
AnswerThe given quadric equation is $(k+1) x^2+2(k+3) x+(k+8)=0$, and roots are real and equal
Then find the value of $k$.
Here, $a=(k+1), b=2(k+3)$ and $c=k+8$
As we know that $d=b^2-4 a c$
Putting the value of $a=(k+1), b=2(k+3)$ and $c=k+8$
$=(2(k+3))^2-4 \times(k+1) \times(k+8)$
$=\left(4 k^2+24 k+36\right)-4\left(k^2+9 k+8\right)$
$=4 k^2+24 k+36-4 k^2-36 k-32$
$=-12 k+4$
The given equation will have real and equal roots, if $D=0$
$-12 k+4=0$
$k=\frac{4}{12}$
$k=\frac{1}{3}$
Therefore, the value of $k =\frac{1}{3}$
View full question & answer→Question 293 Marks
In the following, find the value of k for which the given value is a solution of the given equation:
$\text{kx}^2+\sqrt{2}\text{x}-4=0,\text{x}=\sqrt{2}$
Answer$\text{kx}^2+\sqrt{2}\text{x}-4=0,\text{x}=\sqrt{2}$
Given that $\text{x}=\sqrt{2}$ is a root at the given equation
$\text{kx}^2+\sqrt{2}\text{x}-4=0$
$\Rightarrow\text{x}=\sqrt{2}$ Satisfies the equation
i.e., $\text{k}(\sqrt{2})^2+\sqrt{2}(\sqrt{2})-4=0$
$\Rightarrow2\text{k}+2-4=0$
$\Rightarrow2\text{k}-2=0$
$\Rightarrow2\text{k}=2$
$\Rightarrow\text{k}=1$
View full question & answer→Question 303 Marks
The height of a right triangle is $7\ cm$ less than its base. If the hypotenuse is $13\ cm$, form the quadratic equation to find the base of the triangle.
AnswerGiven that in a right triangle is $7\ cm$ less than its base
Let base of the triangle be denoted by $x$
$\Rightarrow$ Height of the triangle $=(x-7) cm$
We have hypotenuse of the triangle $=13 cm$
We know taht, in a right triangle
$\text { (base }^2+(\text { Height })^2=(\text { Hypotenuse })^2$
$\Rightarrow(x)^2+(x-7)^2=(13)^2$
$\Rightarrow x^2+x^2-14 x+49=169$
$\Rightarrow 2 x^2-14 x+49-169=0$
$\Rightarrow 2 x^2-14 x-120=0$
$\Rightarrow 2\left(x^2-7 x-60\right)=0$
$\Rightarrow x^2-7 x-60=0$
$\therefore$ The required quadratic equation is $x^2-7 x-60=0$
View full question & answer→Question 313 Marks
Ashu is $x$ years old while his mother Mrs. Veena is $x^2$ years old. Five years hence Mrs. Veena will be three times old as Ashu. Find their present ages.
AnswerGiven that, Ashu is $x$ years old while his mother Mrs. Veena is $x^2$ years old.
Ashu's present age $=x$ years and Mrs. Veena's present age $=x^2$ years
And also given that, after $5$ years Mrs. Veena will be three times old as Ashu.
Ashu's age after $5$ years $=(x+5)$ years
And mrs. veena's age after 5 years $=\left(x^2+5\right)$ years
But given that,
$\Rightarrow\left(x^2+5\right)=3(x+5)$
$\Rightarrow x^2+5=3 x+15$
$\Rightarrow x^2-3 x-10=0$
$\Rightarrow x^2-5 x+2 x-10=0$
$\Rightarrow x(x-5)+2(x-5)=0$
$\Rightarrow(x-5)(x+2)=0$
$\Rightarrow x=5 \text { or } x=-2$
Since, age cannot be in negative values. So $x=5$ years.
$\therefore$ Present age of Ashu is $x=5$ years and Present age of Mrs. Veena is $x^2=5^2$ years
$\Rightarrow 25$ years.
View full question & answer→Question 323 Marks
Two number differ by $3$ and their product is $504.$ Find the numbers.
AnswerLet the two numbers be $x$ and $x - 3$ given that $x(x - 3) = 504$
$\Rightarrow x^2 - 3x - 504 = 0$
$\Rightarrow x^2 - 24x + 21x - 504 = 0$
$\Rightarrow x(x - 24) + 21(x - 24) = 0$
$\Rightarrow (x - 24) + 21(x - 24) = 0$
$\Rightarrow (x - 24)(x + 21) = 0$
$\Rightarrow x = 24$ or $x = 21$
Case I: $If x - 3 = 24 - 3 = 21$
Case II: $If x = 21, x = 3 = 24$
$\therefore$ The two numbers are $21, 24$ or $-21, -24$
View full question & answer→Question 333 Marks
The sum of two numbers is $8$ and $15$ times the sum of their reciprocals is also $8$. Find the numbers.
AnswerSum of two nubers $=8$
Let first number $= x$
Then second number $=8- x$
According to the condition,
$15\left[\frac{1}{x}+\frac{1}{8-x}\right]=8$
$\Rightarrow \frac{1}{x}+\frac{1}{8-x}=\frac{8}{15}$
$\Rightarrow \frac{8-x+x}{x(8-x)}=\frac{8}{15}$
$\Rightarrow \frac{8}{x(8-x)}=\frac{8}{15}$
$\Rightarrow x(8-x)=15$
$\Rightarrow 8 x-x^2-15=0$
$\Rightarrow-x^2+8 x-15=0$
$\Rightarrow x^2-8 x+15=0$
$\Rightarrow x^2-3 x-5 x+15=0 $
$\left\{\begin{array}{c} \because 15=-3 \times(-5) \\ -8=-3-5 \end{array}\right\}$
$\Rightarrow x(x-3)-5(x-3)=0$
$\Rightarrow(x-3)(x-5)=0$
Either $x-3=0$, then $x=3$
Or $x-5=0$, then $x=5$
$i.$ If $x=3$, then First number $=3$ and second number $=8-3=5$
$ii.$ If $x=5$, then First number $=5$ and second number $=8-5=3$
Numbers are $3,5$
View full question & answer→Question 343 Marks
The product of Shikha's age five years ago and her age $8$ years later is $30$, her age at both times being given in years. Find her present age.
AnswerLet present age of Shikha = $x$ years
5 years ago, her age was = $(x - 5)$ years
and 8 years later, her age will be = $(x + 8)$ years
According to the condition,
$(x - 5)(x + 8) = 30$
$\Rightarrow x^2 + 3x - 40 = 30$
$\Rightarrow x^2 + 3x - 40 - 30 = 0$
$\Rightarrow x^2 + 3x - 70 = 0$
$\Rightarrow x^2 + 10x - 7x - 70 = 0$
$\Rightarrow x(x + 10) - 7(x + 10) = 0$
$\Rightarrow (x + 10)(x - 7) = 0$
Either $x + 1 0 = 0$, then $x = -10$ which is not possible being negative
Or $x - 7 = 0,$ then $x = 7$
Her present age =$ 7$ years.
View full question & answer→Question 353 Marks
Solve the following quadratic equation by factorization:
$a^2b^2x^2 + b^2x - a^2x - 1 = 0$
AnswerWe have
$a^2b^2x^2 + b^2x - a^2x - 1 = 0$
$[-1 \times a^2b^2 = -a^2b^2 \Rightarrow -a^2b^2 = -a^2 \times b^2 = -a^2 \times b^2]$
$\Rightarrow a^2b^2x^2 + b^2x - a^2x - 1 = 0$
$\Rightarrow b^2x(a^2x + 1) - 1(a^2x + 1) = 0$
$\Rightarrow (a^2x + 1)(b^2x - 1) = 0$
$\Rightarrow a^2x + 1 = 0$ or $b^2x - 1 = 0$
$\Rightarrow\text{x}= -\frac{1}{\text{a}^2}$ and $\text{x}= \frac{1}{\text{b}^2}$ are the two root of the given equation.
View full question & answer→Question 363 Marks
The sum of the squares of two consecutive even numbers is $340$. Find the numbers.
AnswerLet first even number $= 2x$
Then second number $= 2x + 2$
$\Rightarrow (2x)^2 + (2x + 2)^2 = 340$
$\Rightarrow 4x^2 + 4x^2 + 8x + 4 - 340 = 0$
$\Rightarrow 8x^2 + 8x - 336 = 0$
$\Rightarrow x^2+ x - 42 = 0$ (Dividing by $8$)
$\Rightarrow x^2 + 7x - 6x - 42 = 0$
$\Rightarrow x(x + 7) - 6(x + 7) = 0$
$\Rightarrow (x + 7)(x - 6) = 0$
Either $x + 7 = 0$, then $x = -7$
but not possible being negative
or $x - 6 = 0$, then $x = 6$
Number are $12, 14$
View full question & answer→Question 373 Marks
Solve the following quadratic equations by factorization:
$2x^2 + ax - a^2 = 0$
Answer$2x^2 + ax - a^2 = 02x^2 + 2ax - ax - a^2 = 0$
$2x(x + a) - a(x + a) = 0$
$(x + a)(2x - a) = 0$
$x + a = 0$ or $2x - a = 0$
$x = -a$ or $\text{x}=\frac{\text{a}}{2}$
Alternate Answer
First calculate $D = b^2 - 4ac$
Then apply $\text{x} = {-\text{b} \pm \sqrt{\text{D}} \over 2\text{a}}$
We get $x = -a$, $\text{x}=\frac{\text{a}}{2}$
View full question & answer→Question 383 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$3\text{a}^2\text{x}^2+8\text{abx}+4\text{b}^2=0,$ $\text{a}\neq0$
Answer$3\text{a}^2\text{x}^2+8\text{abx}+4\text{b}^2=0,$ $\text{a}\neq0$The given equation in the form of $ax^2 + bx + c = 0$
Here, $a = 3a^2, b = 8ab, c = 4b^2$ $[$given $\text{a}\neq0]$
$D = b^2 - 4ac$
$= (8ab)^2 - 4 \times 3a^2 \times 4b^2$
$= 64a^2b^2 - 48a^2b^2$
$= 16a^2b^2 > 0$
As $Q = 0$, the given equation has real roots, given by
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=-\frac{(8\text{ab})+\sqrt{16\text{a}^2\text{b}^2}}{2\times3\text{a}^2}$
$=\frac{-2\text{b}}{3\text{a}}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=-\frac{(8\text{ab})-\sqrt{16\text{a}^2\text{b}^2}}{2\times3\text{a}^2}$
$=\frac{-2\text{b}}{\text{a}}$
$\therefore$ The roots of the given equation are $\frac{-2\text{b}}{3\text{a}},\frac{-2\text{b}}{\text{a}}$
View full question & answer→Question 393 Marks
Find the value of k for which root are real and equal in the following equations:
$9x^2 - 24x + k = 0$
AnswerThe given equation is $9x^2 - 24x + k = 0$
This equation is in form of $ax^2 + bx + c = 0$
Here, $a = 9, b = -24$ and $c = k$
Given that, the nature of the roots of this equation is real and equal
i.e., $D = b^2 - 4ac = 0$
$\Rightarrow (-24)^2 - 4 \times 9 \times k = 0$
$\Rightarrow 576 - 36k = 0$
$\Rightarrow\text{k}=\frac{576}{36}$
$\therefore$ $k = 16$
$\therefore$ The value of $k = 16$
View full question & answer→Question 403 Marks
In the following, find the value of k for which the given value is a solution of the given equation:
$7\text{x}^2+\text{kx}-3=0,$ $\text{x}=\frac{2}{3}$
AnswerWe are given here that,
$7\text{x}^2+\text{kx}-3=0,$ $\text{x}=\frac{2}{3}$
Now, as we know that $\text{x}=\frac{2}{3}$ is a solution of the quadratic equation, hence it should satify the equation. Therefore substituting $\text{x}=\frac{2}{3}$ in the above equation gives us,
$7\Big(\frac{2}{3}\Big)^2+\text{k}\Big(\frac{2}{3}\Big)-3=0$
$\frac{28+6\text{k}-27}{9}=0$
$6\text{k}=-1$
$\text{k}=-\frac{1}{6}$
Hence, the value of $\text{k}=-\frac{1}{6}$
View full question & answer→Question 413 Marks
A fast train takes one hour less than a slow train for a journey of $200\ km$. If the speed of the slow train is $10\ km/hr$ less than that of the fast train,
find the speed of the two trains.
AnswerTotal journey $= 200\ km$
Let the speed of fast train $= x\ km/hr$
Then speed of slow train $= (x - 10)\ km/hr$
According to the condition,
$\frac{200}{\text{x}-10}-\frac{200}{\text{x}}=1$
$\Rightarrow\frac{200\text{x}-200\text{x}+2000}{\text{x}(\text{x}-10)}=1$
$\Rightarrow\frac{2000}{\text{x}^2-10\text{x}}=1$
$\Rightarrow x^2 - 10x = 2000$
$\Rightarrow x^2 - 10x - 2000 = 0$
$\Rightarrow x^2 - 50x + 40x - 2000 = 0$
$\begin{cases}\because-2000=-50\times40\\-10=-50+40\end{cases}$
$\Rightarrow x(x - 50) + 40 (x - 50) = 0$
$\Rightarrow (x - 50) (x + 40) = 0$
Either $x - 50 = 0$, then $x = 50$
or $x + 40 = 0$, then $x = -40$ but it is not possible being negative
Speed of the fast trin $= 50\ km/hr$
and speed of the slow trin $= 50 - 10 = 40km/hr$
View full question & answer→Question 423 Marks
In the following, determine the set of values of k for which the given quadratic equation has real root:
$3x^2 + 2x + k = 0$
Answer$3x^2 + 2x + k = 0$
Here $a = 3, b = 2, c = k$
$\therefore$ Discriminant $(D) = b^2 - 4ac$
$= (2)^2 - 4 \times 3 \times k$
$= 4 - 12k$
$\because$ Roots are real
$\therefore\text{D}\geq0$
$\Rightarrow4-12\text{k}\geq0$
$\Rightarrow4\geq12\text{k}$
$\Rightarrow12\text{k}\leq4$
$\Rightarrow\text{k}\leq\frac{4}{12}$
$\Rightarrow\text{k}\leq\frac{1}{3}$
View full question & answer→Question 433 Marks
In the following, determine the set of values of $k$ for which the given quadratic equation has real root:
$kx^2 + 6x + 1 = 0$
AnswerThe given equation is $kx^2 + 6x + 1 = 0$, and roots are real.
Then find the value of $k$.
Here, $a = k, b = 6$ and $c = 1$
As we know that $D = b^2 - 4ac$
Putting the value of $a = k, b = 6$ and $c = 1$
$= (6)^2 - 4 \times k \times 1$
$= 36 - 4k$
The given equation will have real roots, if $\text{D}\geq0$
$36-4\text{k}\geq0$
$4\text{k}\leq36$
$\text{k}\leq\frac{36}{4}$
$\text{k}\leq9$
Therefore, the value of $\text{k}\leq9$
View full question & answer→Question 443 Marks
Solve the following quadratic equations by factorization:
$48x^2 - 13x - 1 = 0$
Answer$48x^2 - 13x - 1 = 0$
$\Rightarrow 48x^2 - 16x + 3x - 1 = 0$
$\begin{cases}\because48\times(-1)=48\\\therefore-48=-16\times3\\-13=-16+3\end{cases}$
$\Rightarrow 16x(3x - 1) + 1(3x - 1) = 0$
$\Rightarrow (3x - 1)(16x + 1) = 0$
Either $3x - 1 = 0,$
Then $3x = 1$
$\Rightarrow\text{x}=\frac{1}{3}$
Or $16x + 1 = 0,$
Then $16x = -1$
$\Rightarrow\text{x}=\frac{-1}{16}$
Roots are $\text{x}=\frac{1}{3},\frac{-1}{16}$
View full question & answer→Question 453 Marks
The sum of a number and its positive square root is $\frac{6}{25}.$ Find the number.
AnswerLet the number be $x$
By the hypothesis, we have
$\Rightarrow\text{x}+\sqrt{\text{x}}=\frac{6}{25}$
Let us assume that $x = y^2$, we get
$\Rightarrow\text{y}^2+\text{y}=\frac{6}{25}$
$\Rightarrow25\text{y}^2+25\text{y}-6 = 0$
The value of 'y' can be obtaines by $\text{y} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
Where $a = 25, b = 25, c = -6$
$\Rightarrow\text{y}=\frac{-25\pm\sqrt{625+600}}{50}$
$\Rightarrow\text{y}=\frac{-25\pm35}{50}$
$\Rightarrow\text{y}=\frac{1}{5}$ or $\frac{-11}{10}$
$\Rightarrow\text{x}=\text{y}^2$
$\Rightarrow\text{x}=\Big(\frac{1}{5}\Big)^2$
$\Rightarrow\text{x}=\frac{1}{25}$
$\therefore$ The number $\text{x}=\frac{1}{25}$
View full question & answer→Question 463 Marks
In the following, determine whether the given quadratic equation have real roots and if so, find the roots:
$\sqrt{3}\text{x}^2+10\text{x}-8\sqrt{3}=0$
Answer$\sqrt{3}\text{x}^2+10\text{x}-8\sqrt{3}=0$
The given equation is in the form of $ax^2 + bx + c = 0$
Here $\text{a}=\sqrt{3},\text{b}=10$ and $\text{c}=-8\sqrt{3}$
$\text{D}=\text{b}^2-4\text{ac}$
$=(10)^2-4\times\sqrt{3}\times-8\sqrt{3}$
$=100+96=196$
AS $Q > 0$, the given equation has real roota, given by
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-10+\sqrt{196}}{2\times\sqrt{3}}$
$=\frac{2\sqrt{3}}{3}=\frac{2}{\sqrt{3}}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-10-\sqrt{196}}{2\times\sqrt{3}}$
$=-4\sqrt{3}$
$\therefore$ The roots of the equation are $\frac{2}{\sqrt{3}}$ and $-4\sqrt{3}$
View full question & answer→Question 473 Marks
Find the values of k for which the given quadratic equation has real and distinct root:
$x^2 - kx + 9 = 0$
AnswerThe given quadric equation is $x^2 - kx + 9 = 0$, and roots are real and distinct.
Then find the value of k.
$A = 1, b = -k$ and $c = 9$
As we know that $D = b^2 - 4ac$
Putting the value of $a = 1, b = -k$ and $c = 9$
$\Rightarrow D = (k)^2 - 4 \times 1 \times 9$
$\Rightarrow D = K^2 - 36$
The given equation will have real and distinct roots, if $D > 0$
$\Rightarrow k^2 - 36 > 0$
Now factorizing of the above equation
$\Rightarrow k^2 - 36 > 0$
$\Rightarrow k^2 > 36$
$\Rightarrow\text{k}>\sqrt{36}$
$\Rightarrow\text{k}=\pm6$
$\Rightarrow k < -6 or k > 6$
Therefore, the value of $k < -6$ or,$ k > 6$
View full question & answer→Question 483 Marks
Is there any value of $'a'$ for which the equation $x^2 + 2x + (a^2 + 1) = 0$ has real root?
AnswerLet quadratic equation $x^2 + 2x + (a^2 + 1) = 0$
Here $a = 1, b = 2$ and $c = (a^2 + 1)$
As we know that $D = b^2 - 4ac$
Putting the value of $a = 1, b = 2$ and c = (a^2 + 1)
$\Rightarrow D = (2)^2 - 4 \times 1 \times (a^2 + 1)$
$\Rightarrow D = 4 - 4(a^2 + 1)$
$\Rightarrow D = -4a^2$
The given equation will have equal roots, id $D > 0$
$i.e., -4a^2 > 0$
$\Rightarrow a^2 < 0$
Which is not possible, as the square of any number is always positive.
Thus, No, there is no any real value of a which the given equation has real roots.
View full question & answer→Question 493 Marks
Solve the following quadratic equations by factorization:
$\text{x}^2+\Big(\text{a}+\frac{1}{\text{a}}\Big)\text{x}+1=0$
AnswerWe have been given
$\text{x}^2+\Big(\text{a}+\frac{1}{\text{a}}\Big)\text{x}+1=0$
Therefore,
$\text{x}^2+\text{ax}+\frac{1}{\text{a}}\text{x}+1=0$
$\text{x}(\text{x}+\text{a})+\frac{1}{\text{a}}(\text{x}+\text{a})=0$
$\Big(\text{x}+\frac{1}{\text{a}}\Big)(\text{x}+\text{a})=0$
Therefore,
$\text{x}+\frac{1}{\text{a}}=0$
$\text{x}=-\frac{1}{\text{a}}$
or, $\text{x}+\text{a}=0$
$\text{x}=-\text{a}$
Hence, $\text{x}=-\frac{1}{\text{a}}$ or $\text{x}=-\text{a}$
View full question & answer→Question 503 Marks
Solve the following quadratic equations by factorization:
$\frac{1}{\text{x}}-\frac{1}{\text{x}-2}=3,$ $\text{x}\neq0,2$
Answer$\frac{1}{\text{x}}-\frac{1}{\text{x}-2}=3$ $(\text{x}\neq0,2)$
$\frac{\text{x}-2-\text{x}}{\text{x}(\text{x}-2)}=3$
$\Rightarrow\frac{-2}{\text{x}^2-2\text{x}}=3$
$\Rightarrow 3x^2 - 6x = -2$
$\Rightarrow 3x^2 - 6x + 2 = 0$
Here $a = 3, b = -6, c = 2$
$D = b^2 - 4ac = (-6)^2 - 4 \times 3 \times 2$
$= 36 - 24 = 12$
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$\text{x}={-(-6) \pm \sqrt{12} \over 2\times3}$
$\text{x}={6 \pm2 \sqrt{3} \over6}$
$\text{x}=\frac{2(3\pm\sqrt{3})}{6}$
$\text{x}=\frac{3\pm\sqrt{3}}{3}$
View full question & answer→