Question 513 Marks
Find the value of k for which the following equations have real and equal roots:
$(k + 1)x^2 - 2(k - 1)x + 1 = 0$
Answer$(k+1) x^2-2(k-1) x+1=0$
Here $a=k+1, b=-2(k-1)$ and $c=1$
$\therefore$ Discriminant $(D) = b^2 - 4ac$
$= [-2(k - 1)]^2 - 4(k + 1) \times 1$
$= 4(k^2 - 2k + 1) - 4(k + 1)$
$= 4k^2 - 8k + 4 - 4k - 4$
$= 4k^2 - 12k$
$\because$ Roots are real and equal
$\therefore$ $D = 0$
$4k^2- 12k = 0$
$k^2 - 3k = 0$ (Dividing by $4$)
$k(k - 3) = 0$
Either$ k = 0$
$or k - 3 = 0$, then$ k = 3$
$\therefore$ $k = 0, 3$
View full question & answer→Question 523 Marks
Find the value of k for which the root are real and equal in the following equations:
$kx^2 + 4x + 1 = 0$
Answer$kx^2 + 4x + 1 = 0$
Here $a = k, b = 4, c = 1$
$\therefore$ Discriminant $(D) = b^2 - 4ac = (4)^2 - 4 \times k \times 1 = 16 - 4k$
$\because$ Roots are real and equal
$\therefore$ $D = 0$
$\Rightarrow 16 - 4k = 0$
$\Rightarrow 4k = 16$
$\Rightarrow\text{k}=\frac{16}{4}=4$
Hence, $k = 4$
View full question & answer→Question 533 Marks
The sum of a number and its reciprocal is $\frac{17}{4}.$ Find the number.
AnswerLet a number be x abd its reciprocal is $\frac{1}{\text{x}}$
Then according to question
$\text{x}+\frac{1}{\text{x}}=\frac{17}{4}$
$\frac{\text{x}^2+1}{\text{x}}=\frac{17}{4}$
By cross multoplication,
$4x^2 + 4 = 17x$
$4x^2 - 17x + 4 = 0$
$4x^2 - 17x + 4 = 0$
$4x^2 - x - 16x + 4 = 0$
$x(4x - 1) - 4(4x - 1) = 0$
$(4x - 1)(x - 4) = 0$
$(4x - 1) = 0$
$\text{x}=\frac{1}{4}$
Or $(x - 4) = 0$
$x = 4$
Thus, two consecutive number be either $4$ or $\frac{1}{4}$
View full question & answer→Question 543 Marks
Solve the following quadratic equations by factorization: $x^2-x-a(a+1)=0$
Answer$x^2 - x - a(a + 1) = 0$
$\Rightarrow x^2 + {(a) - (a + 1)}x - a(a + 1) = 0$
${ 1 = a - a + 1}$
$\Rightarrow x^2 + ax - (a + 1)x - a(a + 1) = 0$
$\Rightarrow x(x + a) - (a + 1)(x + a) = 0$
$\Rightarrow (x + a)(x - a - 1) = 0$
Either $x+a=0$, then $x=-a$
or $x-a-1=0$, then $x=a+1$
$\therefore$ Roots are $-a, a+1$
View full question & answer→Question 553 Marks
Solve the following quadratic equations by factorization:
$5x^2 - 3x - 2 = 0$
AnswerWe have,$5x^2 - 3x - 2 = 0$
$\Rightarrow 5x^2 - 5x + 2x - 2 = 0$
$\Rightarrow 5x(x - 1) + 2(x - 1) = 0$
$\Rightarrow (x - 1)(5x + 2) = 0$
$\Rightarrow (x - 1) = 0$ or $5x + 2 = 0$
$\Rightarrow x = 1$ or $\text{x}=-\frac{2}{5}$
$\therefore$ x = 1 and $\text{x}=-\frac{2}{5}$ are the two roots of the given equation.
View full question & answer→Question 563 Marks
The product of Ramu's age (in years) five years ago and his age (in years) nine years later is $15$. Determine Ramu's present age.
AnswerLet the present age of Ramu be a x years
Given that,
The product of his age years ago and his age y nine years later is 15.
Now, Ramu's age five years ago $= (x - 5)$ years
And Ramu's age nine years later $= (x + 9)$ years
Given that,
$(x - 5)(x + 9) = 15$
$\Rightarrow x^2 + 9x - 5x - 45 = 15$
$\Rightarrow x^2 + 4x - 60 = 0$
$\Rightarrow x^2 + 10x - 6x - 60 = 0$
$\Rightarrow x(x + 10) - 6(x + 10) = 0$
$\Rightarrow (x + 10)(x - 6) = 0$
$\Rightarrow x + 10 = 0 or x - 6 = 0$
$\Rightarrow x = -10 or x = 6$
Since, age cannot be in negative value , so $x = 6$ years
$\therefore$ The present age of Ramu is $6$ years.
View full question & answer→Question 573 Marks
Find two consecutive odd positive integers, sum of whose squares is $970$.
AnswerLet two consecutive positive integers be x and $x + 2$
$A.T.Q., (x)^2 + (x + 2)^2 = 970$
$\Rightarrow x^2 + x^2 + 4x + 4 - 970 = 0$
$\Rightarrow 2x^2 + 4x - 966 = 0$
$\Rightarrow x^2 + 2x - 483 = 0$
$\Rightarrow x^2 + 23x - 21x - 483 = 0$
$\Rightarrow x(x + 23) - 21(x + 23) = 0$
$\Rightarrow (x - 21)(x + 23) = 0$
Either $x - 21 = 0 or x + 23 = 0$
$x = 21 or x = -23$ (rejected being -ve)
As integers should be +ve
$x = 21$ and $x + 2 = 21 + 2 = 23$
Hence integers are $21, 23$
View full question & answer→Question 583 Marks
Solve the following quadratic equations by factorization:
$\text{x}-\frac{1}{\text{x}}=3,\text{x}\neq0$
Answer$\text{x}-\frac{1}{\text{x}}=3$
$\Rightarrow\text{x}^2-3\text{x}-1=0$
Here $\text{a}=1,\text{b}=-3,\text{c}=-1$
$\therefore\text{D}=\text{b}^2-4\text{ac}$
$=(-3)^2-4\times(1)(-1)$
$=9+4=13$
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$\text{x}={-(-3) \pm \sqrt{13} \over 2\times1}$
$\text{x}=\frac{3\pm\sqrt{13}}{2}$
View full question & answer→Question 593 Marks
The sum of the squares of two consecutive multiples of $7$ is $637$. Find the multiples.
AnswerLet one of the number be 7x then the other number be $7(x + 1)$
Then according to question,
$\Rightarrow (7x)^2 + [7(x + 1)]^2 = 637$
$\Rightarrow 49x^2 + 49(x^2 + 2x + 1) = 637$
$\Rightarrow 49x^2 + 49x^2 + 98x + 49 - 637 = 0$
$\Rightarrow 98x^2 + 98x - 588 = 0$
$\Rightarrow x^2 + x - 6 = 0$
$\Rightarrow x^2 + 3x - 2x - 6 = 0$
$\Rightarrow x(x + 3) - 2(x + 3) = 0$
$\Rightarrow (x - 2)(x + 3) = 0$
$\Rightarrow x - 2 = 0 or x + 3 = 0$
$\Rightarrow x = 2 or x = -3$
Since, the numbers are mutiples of 7,
Therefore, one number $= 7 × 2 = 14$
Then another number will be $7(x + 1) = 7 × 3 = 21$
Thus, the two consecutive multiples of $7$ are $14$ and $21$
View full question & answer→Question 603 Marks
The product of two successive integral multiples of $5$ is $300$. Determine the multiples.
AnswerGiven that the product of two successive integral multiples of $5$ is $300$
Let the integers be $5x$, and $5(x + 1)$
Then, ny the integers be $5x$ and $5(5x + 1)$
Then, by the hypothesis, we have
$5x \times 5(5x + 1) = 300$
$\Rightarrow 25x(x + 1) = 300$
$\Rightarrow x^2 + x = 12$
$\Rightarrow x^2 + x - 12 = 0$
$\Rightarrow x^2 + 4x - 3x - 12 = 0$
$\Rightarrow x(x + 4) - 3(x + 4) = 0$
$\Rightarrow (x + 4)(x - 3) = 0$
$\Rightarrow x = -4 or x = 3$
$If x = -4, 5x = -20, 5(x + 1) = -15$
$x = 3, 5x = 15, 5(x + 1) = 20$
The two successive integral multiples are $15, 20$ or $-15, -20$
View full question & answer→Question 613 Marks
Find the roots of the following quadratic equation (if they exist) by the method of completing the square.
$3\text{x}^2+11\text{x}+10=10$
AnswerWe have been given that,
$3\text{x}^2+11\text{x}+10=0$
Now, divide throughout by 3 we get,
$\text{x}^2+\frac{11}{3}\text{x}+\frac{10}{3}=0$
Now take the constant term to the R.H.S. and we get,
$\text{x}^2+\frac{11}{3}\text{x}=-\frac{10}{3}$
Now add square of half co-efficient of 'x' on both the sides. we have,
$\text{x}^2+\frac{11}{3}\text{x}+\Big(\frac{11}{6}\Big)^2=\Big(\frac{11}{6}\Big)^2-\frac{10}{3}$
$\text{x}^2+\Big(\frac{11}{6}\Big)^2+2\Big(\frac{11}{3}\Big)\text{x}=\frac{1}{36}$
$\Big(\text{x}^2+\frac{11}{6}\Big)^2=\frac{1}{36}$
Since R.H.S. is a positive number, therefore the roots of the equation exist.
So, now take the square root on both the sides and we get,
$\text{x}+\frac{11}{6}=\pm\frac{1}{6}$
$\text{x}=-\frac{11}{6}\pm\frac{1}{6}$
Now, we have the values of 'x' as
$\text{x}=-\frac{11}{6}\pm\frac{1}{6}$
$\text{x}=-\frac{5}{3}$
Also we have,
$\text{x}=-\frac{11}{6}-\frac{1}{6}$
$\text{x}=-2$
Therefore the roots of the equation are -2 and $-\frac{5}{3}$
View full question & answer→Question 623 Marks
In the following, determine whether the given values are solution of the given equation or not:
$x^2 - 3x + 2 = 0, x = 2, x = -1$
Answer$x^2 - 3x + 2 = 0. x = 2, x = -1$
$$Here, $L.H.S. = x^2 - 3x + 2$ and $R.H.S. = 0$
Now, substitute $x = 2$ in $L.H.S.$
We get $(2)^2 - 3(2) + 2 = 4 - 6 + 2$
$= 6 - 6$
$= 0$
R.H.S.
Since, L.H.S. = R.H.S.
x = 2 is a solution for the given equation.
Similarly,
Now substitute x = -1 in L.H.S.
We get $(-1)^2 - 3(-1) + 2$
1 + 3 + 2 = 6 $\neq$ R.H.S.
Since L.H.S $\neq$ R.H.S.
x = -1 is not a solution for thr given equatoion.
View full question & answer→Question 633 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$2\text{x}^2+5\sqrt{3}\text{x}+6=0$
Answer$2\text{x}^2+5\sqrt{3}\text{x}+6=0$
Here, $\text{a}=2,\text{b}=5\sqrt{3},\text{c}=6$
$\therefore$ Discriminant $(\text{D})=\text{b}^2-4\text{ac}$
$=(5\sqrt{3})^2-4\times2\times6$
$=75-48=27$
$\because\text{D}>0$
$\therefore$ Roots are real and unequal
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$=\frac{-5\sqrt{3}\pm\sqrt{27}}{2\times2}$
$=\frac{-5\sqrt{3}\pm3\sqrt{3}}{4}$
$\therefore\text{x}=\frac{-5\sqrt{3}+3\sqrt{3}}{4}=\frac{-2\sqrt{3}}{4}$
$\text{x}=\frac{-\sqrt{3}}{2}$
and $\text{x}=\frac{-5\sqrt{3}-3\sqrt{3}}{4}=\frac{-8\sqrt{3}}{4}$
$\text{x}=-2\sqrt{3}$
Roots are $\frac{-\sqrt{3}}{2},-2\sqrt{3}$
View full question & answer→Question 643 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$2\text{x}^2-2\sqrt{2}\text{x}+1=0$
Answer$2\text{x}^2-2\sqrt{2}\text{x}+1=0$
$\Rightarrow(\sqrt{2}\text{x})^2-2\times\sqrt{2}\text{x}\times1+(1)^2=0$
$\Rightarrow(\sqrt{2}\text{x}-1)^2=0$
$\Rightarrow\sqrt{2}\text{x}-1=0$
$\Rightarrow\sqrt{2}\text{x}=1$
$\Rightarrow\text{x}=\frac{1}{\sqrt{2}}$
$\therefore\text{x}=\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}$
View full question & answer→Question 653 Marks
Find the value of k for which root are real and equal in the following equations:
$2kx^2 - 40x + 25 = 0$
AnswerThe given quadric equation is $2kx^2 - 40x + 25 = 0$, and roots real and equal
Then find the value of $k$.
Here, $a = 2k, b = -40$ and $c = 25$
As we know that $D = b^2 - 4ac$
Putting the value of $a = 2k, b = -40$ and $c = 25$
$= (-40)^2 - 4 \times 2k \times 25$
$= 1600 - 200k$
The given equation will have real and equal roots , if $D = 0$
$Thus, 1600 - 200k = 0$
$200k = 1600$
$\text{k}=\frac{1600}{200}$
$k = 8$
Therefore, the value of $k = 8$
View full question & answer→Question 663 Marks
John and javanti together have $45$ marbles. Both of them lost $5$ marbles each, and the product of the number of marber of marbles they now have is $128$. From the quadratic equation to find how many marbles had to start with, if john and x marbles.
AnswerIt is given that John had ' $x$ ' marbles.
We are also given that both John and javanti had 45 marbles together.
So, Javanti should have ' 45 - x' marbles with her.
Now, it is given taht both of them lose 5 marbles each.
So, in the new situation, John will have ' $x -5$ ' marbles and Javanti will have ' $45- x -5$ ' marbles.
Also it is given taht the product of the number of marbles pof marbles both of them now is $128$ .
Therefore,
$(x - 5)(45 - x - 5) = 128$
$(x - 5)(40 - x) = 128$
$40x - x^2 - 200 + 5x - 128 = 0$
$x^2 - 45x + 200 + 128 = 0$
$x^2 - 45x + 328 = 0$
Hence, this is the required quadratic equation.
View full question & answer→Question 673 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$3\text{x}^2+2\sqrt{5}\text{x}-5=0$
Answer$3\text{x}^2+2\sqrt{5}\text{x}-5=0$
Here $\text{a}=3,\text{b}=2\sqrt{5},\text{c}=-5$
$\therefore$ Discriminant $(\text{D})=\text{b}^2-4\text{ac}$
$=(2\sqrt{5})^2-4\times3\times(-5)$
$=20+60=80$
$\because\text{D > 0}$
$\therefore$ Roots are real and unequal
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$=\frac{-2\sqrt{5}\pm\sqrt{80}}{2\times3}$
$=\frac{-2\sqrt{5}\pm4\sqrt{5}}{6}$
$=\frac{-\sqrt{5}\pm2\sqrt{5}}{3}$
$\therefore\text{x}=\frac{-\sqrt{5}+2\sqrt{5}}{3}$
$\text{x}=\frac{\sqrt{5}}{3}$
and $\text{x}=\frac{-\sqrt{5}-2\sqrt{5}}{3}$
$=\frac{-3\sqrt{5}}{3}$
$\text{x}=-\sqrt{5}$
Roots are $-\sqrt{5},\frac{\sqrt{5}}{3}$
View full question & answer→Question 683 Marks
Three consecutive positive integers are such that the sum of the square of the first and the product of other two is $46$, find the integers.
AnswerLet first number $= x$
Then second number $= x + 1$
and third number $= x + 2$
According co the condition,
$(x)^2 + (x + 1)(x + 2) = 46$
$\Rightarrow x^2 + x^2 + 3x + 2 = 46$
$\Rightarrow 2x^2 + 3x + 2 - 46 = 0$
$\Rightarrow 2x^2 + 3x - 44 = 0$
$\Rightarrow 2x^2 + 11x - 8x - 44 = 0$
$\Rightarrow x(2x + 11) - 4(2x + 11) = 0$
$\Rightarrow (2x + 11)(x - 4) = 0$
Either $2x + 11 = 0$, then $\text{x}=\frac{-11}{2}$ which is not possible being fraction or $x - 4 = 0$, then $x = 4$
Numbers are $4, 5, 6$
View full question & answer→Question 693 Marks
The sum of two numbers is $18$. The sum of their reciprocals is $\frac{1}{4}$ Find the numbers.
AnswerSum of two numbers $= 18$
Let one number $= x$
Then second number $= 18 - x$
According to the condition,
$\frac{1}{\text{x}}+\frac{1}{18-\text{x}}=\frac{1}{4}$
$\Rightarrow\frac{18-\text{x}+\text{x}}{\text{x}(18-\text{x})}=\frac{1}{4}$
$\Rightarrow\frac{18}{18\text{x}-\text{x}^2}=\frac{1}{4}$
$72 = 18x - x^2$
$x^2 - 18x + 72 = 0$
$\Rightarrow x^2 - 12x - 6x + 72 = 0$
$\Rightarrow x(x - 12) - 6(x - 12) = 0$
$\Rightarrow (x - 12)(x - 6) = 0$
Either $x - 12 = 0$, then $x = 12$
or $x - 6 = 0$, then $x = 6$
- If $x = 12$, then
First number = 12 and second number $= 18 - 12 = 6$
- If $x = 6$, then
First number = 6 then second number $= 18 - 6 = 12$
Number are $6, 12$ View full question & answer→Question 703 Marks
Solve the following quadratic equations by factorization:
$abx^2 + (b^2 - ac)x - bc = 0$
AnswerWe have,
$abx^2 + (b^2 - ac)x - bc = 0$
$[abx - bc = -ab^2c$
$\Rightarrow -ab^2c = b^2 \times -ac and b^2- ac = b^2 + (-ac)]$
$\Rightarrow abx^2 + b^2x - acx - bc = 0$
$\Rightarrow bx(ax + b) - c(ax + b) = 0$
$\Rightarrow (ax + b)(bx - c) = 0$
$\Rightarrow ax + b = 0 or bx - c = 0$
$\Rightarrow\text{x}=-\frac{\text{b}}{\text{a}}$ or $\text{x}=\frac{\text{c}}{\text{b}}$
$\therefore\text{x}=-\frac{\text{b}}{\text{a}}$ and $\text{x}=\frac{\text{c}}{\text{b}}$ are the two roots the given equation.
View full question & answer→Question 713 Marks
Write the value of k for which the quadratic equation $x^2 - kx + 4 = 0$ has equal roots.
AnswerThe given quadric equation is $x^2 - kx + 4 = 0$, and roots are equal.
Then find the value of k.
Here, $a = 1, b = -k$ and $c = 4$
As we know that $D = b^2 - 4ac$
Putting the value of $a = 1, b = -k$ and $c = 4$
$= (-k)^2 - 4 \times 1 \times 4$
$= k2 - 16$
The given equation will have equal roots, if $D = 0$
$k^2 - 16 = 0$
$k^2 = 16$
$\text{k}=\sqrt{16}$
$\text{k}=\pm4$
Therefore, the value of $\text{k}=\pm4$
View full question & answer→Question 723 Marks
The difference of squares of two numbers is $180.$ The square of the smaller number is $8$ times the larger number. Find two numbers.
AnswerLet the number be x
By the given hypothesis, we have
According to question $x^2 - y^2 = 180$ and $y^2 = 8x$
$\Rightarrow x^2 - 8x = 180$
$\Rightarrow x^2 - 8x - 180 = 0$
$\Rightarrow x^2 + 10x - 18x - 180 = 0$
$\Rightarrow x(x + 10) - 18(x + 10) = 0$
$\Rightarrow (x + 10)(x - 18) = 0$
$\Rightarrow x = -10 or x = 18$
Case I: $x = 18$
$\Rightarrow 8x = 8 \times 18 = 144$
Lerger number $=\sqrt{144}=\pm12$
Case II: $x = -10$
Square of larger number 8x = -80 here no perfect square exist, hence the numbers are $18, 12.$
View full question & answer→Question 733 Marks
Solve the following quadratic equations by factorization:
$25x(x + 1) = -4$
AnswerWe have been given
$25x(x + 1) = -4$
$25x^2 + 25x + 4 = 0$
$25x^2 + 20x + 5x + 4 = 0$
$5x(5x + 4) + 1(5x + 4) = 0$
$(5x + 1)(5x + 4) = 0$
Therefore,
$5x + 1 = 0$
$5x = -1$
$\text{x}=\frac{-1}{5}$
Or, $5x + 4 = 0$
$5x = -4$
$\text{x}=\frac{-4}{5}$
Hence, $\text{x}=\frac{-1}{5}$ or $\text{x}=\frac{-4}{5}$
View full question & answer→Question 743 Marks
Solve the following quadratic equations by factorization:
$6x^2 + 11x + 3 = 0$
AnswerWe have been given
$6x^2 + 11x + 3 = 0$
$6x^2 + 9x + 2x + 3 = 0$
$3x(2x + 3) + 1(2x + 3) = 0$
$(2x + 3)(3x + 1) = 0$
$2x + 3 = 0$
$\text{x}=\frac{-3}{2}$
Or, $3x + 1 = 0$
$\text{x}=\frac{-1}{3}$
Hence, $\text{x}=\frac{-3}{2}$ or $\text{x}=\frac{-1}{3}$
View full question & answer→Question 753 Marks
A two-digit number is such that the product of its digits is $8$. When $18$ is subtracted from the number, the digits interchange their places. Find the number.
AnswerLet the two digits be x and x -$ 2$
Given that the product of their digit is $8$
$\Rightarrow x(x - 2) = 8$
$\Rightarrow x^2 - 2x - 8 = 0$
$\Rightarrow x^2 - 4x + 2x - 8 = 0$
$\Rightarrow x(x - 4) + 2(x - 4) = 0$
$\Rightarrow (x - 4)(x + 2) = 0$
$\Rightarrow x = 4 or x = -2$
Considering the positive value $x = 4, x - 2 = 2$
$\therefore$ The two digit number is $42$
View full question & answer→Question 763 Marks
Write the sum of real roots of the equation $x^2 + |x| - 6 = 0$
AnswerThe given quadric equation is $x^2 + |x| - 6 = 0$
Here, $\text{a}=1,\text{b}=\pm1$ and $\text{c}=-6$
As we know that $\text{D}=\text{b}^2-4\text{ac}$
Putting the value of $\text{a}=1,\text{b}=\pm1$ and $\text{c}=-6$
$=(\pm1)^2-4\times1\times-6$
$=1+24$
$=25$
Since, $\text{D}\geq0$
Therefore, root of the given equation are real and distinct.
Thus, sum of the roots be $= 0$
View full question & answer→Question 773 Marks
Solve the following quadratic equations by factorization:
$\text{x}^2-\Big(\sqrt{3}+1\Big)\text{x}+\sqrt{3}=0$
Answer$\text{x}^2-\Big(\sqrt{3}+1\Big)\text{x}+\sqrt{3}=0$$\Rightarrow\text{x}^2-\sqrt{3}\text{x}-\text{x}+\sqrt{3}=0$
$\Rightarrow\text{x}\Big(\text{x}-\sqrt{3}\Big)-1\Big(\text{x}-\sqrt{3}\Big)=0$
$\Rightarrow\Big(\text{x}-\sqrt{3}\Big)(\text{x}-1)=0$
Either $\text{x}-\sqrt{3}=0,$ then $\text{x}=\sqrt{3}$
or $\text{x}-1=0,$ then $\text{x}=1$
$\therefore$ Roots are $\text{x}=\sqrt{3},1$
View full question & answer→Question 783 Marks
Determine the nature of the root of following quadratic equation:
$9a^2b^2x^2 - 24abcdx + 16c^2d^2 = 0$, $\text{a}\neq0,\text{b}\neq0$
AnswerThe given quadric equation is $9a^2b^2x^2 - 24abcdx + 16c^2d^2 = 0$
Here, $a = 9a^2b^2, b = -24$abcd and $c = 16c^2d^2$
As we know that $D = b^2 - 4ac$
Putting the value of $a = 9a^2b^2, b = -24abcd$ and $c = 16c^2d^2$
$= (24abcd)^2 - 4 \times 9a^2b^2 \times 16c^2d^2$
$= (576a^2b^2c^2d^2) - 576a^2b^2c^2d^2$
$= 0$
Since, $D = 0$
Therefore, root of the given equation are real and equal.
View full question & answer→Question 793 Marks
Solve the following quadratic equations by factorization:
$\frac{\text{x}-1}{\text{x}-2}+\frac{\text{x}-3}{\text{x}-4}=3\frac{1}{3},$ $\text{x}\neq2, 4$
AnswerWe have been given
$\frac{\text{x}-1}{\text{x}-2}+\frac{\text{x}-3}{\text{x}-4}=3\frac{1}{3}$
$3(x^2 - 5x + 4 + x^2 - 5x + 6) = 10(x^2 - 6x + 8)$
$4x^2 - 30x + 50 = 0$
$2x^2- 15x + 25 = 0$
$2x^2- 10x - 5x + 25 = 0$
$2x(x - 5) - 5(x - 5) = 0$
$(2x - 5)(x - 5) = 0$
Therefore,
$2x - 5 = 0$
$2x = 5$
$\text{x}=\frac{5}{2}$
or, $x - 5 = 0$
$x = 5$
Hence, $\text{x}=\frac{5}{2}$ or $x = 5$
View full question & answer→Question 803 Marks
The sum of squares of two consecutive odd positive integers is $394$. Find them.
AnswerLet two consecutive odd positive integer be $(2x - 1)$ and other $(2x + 1)$
Then according to question,
$(2x + 1)^2 + (2x - 1)^2 = 394$
$4x^2 + 4x + 1 + 4x^2 - 4x + 1 = 394$
$8x^2 + 2 = 394$
$8x^2 = 394 - 2$
$\text{x}^2=\frac{392}{8}$
$\text{x}^2=49$
$\text{x}=\sqrt{49}$
$\text{x}=\pm7$
Since, x beging a positive numner, so x cannot be negative.
Therefore,
When $x = 7$ then odd positive
$2x - 1 = 2 × 7 - 1$
$2x - 1 = 13$
And, $2x + 1 = 2 × 7 + 1$
$2x + 1 = 15$
Thus, two consecutive odd positive integer be 13, 15
View full question & answer→Question 813 Marks
Since, $\text{x}=-\frac{1}{2},$ is a solution of the quadratic equation $3x^2 + 2kx - 3 = 0$, find the value of k.
AnswerSince, $\text{x}=-\frac{1}{2},$ is a solution of the quadratic equation $3x^2 + 2kx - 3 = 0$
So, it satisfies the given equation,
$\therefore3\Big(-\frac{1}{2}\Big)^2+2\text{k}\Big(-\frac{1}{2}\Big)-3=0$
$\Rightarrow\frac{3}{4}-\text{k}-3=0$
$\Rightarrow\text{k}=\frac{3}{4}-3$
$\Rightarrow\text{k}=\frac{3-12}{4}$
$\Rightarrow\text{k}=-\frac{9}{4}$
Thus, the value k is $-\frac{9}{4}$
View full question & answer→Question 823 Marks
Solve the following quadratic equations by factorization:
$x^2 + 2ab = (2a + b)x$
AnswerWe have
$x^2 + 2ab = (2a + b)x$
$\Rightarrow x^2 - (2a + b)x + 2ab = 0$
[$\because$ $2ab = -8a \times -b$
$\Rightarrow -(8a + b) = -8a - b$
$\Rightarrow x^2 - 2ax - bx + 2ab = 0$
$\Rightarrow x - (x - 8a) - b(x - 2a) = 0$
$\Rightarrow (x - 8a)(x - b) = 0$
$\Rightarrow x - 8a = 0 or x - b = 0$
$\Rightarrow x = 8a = 0$ or $x = b$
$\therefore$ x = 8a and x = b are the two roots of the given equation.
View full question & answer→Question 833 Marks
Solve the following quadratic equations by factorization:
$3x^2 = -11x - 10$
Answer$3x^2 = -11x - 10$
$\Rightarrow 3x^2 + 11x + 10 = 0$
$\Rightarrow 3x^2 + 11x + 10 = 0$
$\begin{cases}\because3\times10=30\\\therefore30=5\times6\\11=5+6\end{cases}$
$\Rightarrow 3x^2 + 6x + 5x + 10 = 0$
$\Rightarrow 3x(x + 2) + 5(x + 2) = 0$
$\Rightarrow (x + 2)(3x + 5) = 0$
Either $x + 2 = 0$, then $x = -2$
Or $3x + 5 = 0.$ then $3x = -5$
$\Rightarrow\text{x}=\frac{-5}{3}$
$\therefore$ Roots are x = -2, $\frac{-5}{3}$
View full question & answer→Question 843 Marks
In the following, determine the set of values of k for which the given quadratic equation has real root:
$2x^2 + 3x + k = 0$
AnswerThe given equation is $2x^2 + 3x + k = 0$
Given that the quadratic equation has real roots i.e.,
$\text{D}=\text{b}^2-4\text{ac}\geq0$
Given here $\text{a}=2,\text{b}=3,\text{c}=\text{k}$
$\Rightarrow9-4\times2\times\text{k}\geq0$
$\Rightarrow9-8\text{k}\geq0$
$\Rightarrow9\geq8\text{k}$
$\Rightarrow\text{k}\leq\frac{9}{8}$
The value of k does not exceed $\frac{4}{8}$ to have roots.
View full question & answer→Question 853 Marks
Find the value of k for which root are real and equal in the following equations:
$4x^2 - 3kx + 1 = 0$
Answer$4x^2 - 3kx + 1 = 0 $Here $a = 4, b = -3k, c = 1$
$\therefore$ Discriminant $(D) = b^2 - 4ac$
$= (-3k)^2 - 4 \times 4 \times 1$
$= 9k^2 - 16$
$\therefore$ Roots are real and equal
$\therefore$ $D = 0$
$\Rightarrow 9k^2 - 16 = 0$
$\Rightarrow 9k^2 = 16$
$\Rightarrow\text{k}^2=\frac{16}{9}$
$\text{k}=\Big(\pm\frac{4}{3}\Big)^2$
$\therefore\text{k}=\pm\frac{4}{3}$
View full question & answer→Question 863 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$16x^2 = 24x + 1$
Answer$16x^2 = 24x + 1$
$\Rightarrow 16x^2 - 24x - 1 = 0$
$\therefore$ Discriminate $= b^2 - 4ac$
$= (-24)^2 - 4 \times 16 \times (-1)$
$= 576 + 64 = 640$
$\because$ D > 0
$\therefore$ Roots are real and unequal
$\therefore\text{x} = {-\text{b}\pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$=\frac{-(-24)\pm\sqrt{640}}{2\times16}$
$=\frac{24\pm8\sqrt{10}}{32}$ (Dividng by 8)
$=\frac{3\pm\sqrt{10}}{4}$
$\therefore$ Roots are $\frac{3+\sqrt{10}}{4},\frac{3-\sqrt{10}}{4}$
View full question & answer→Question 873 Marks
Solve for $x.$
$\frac{1}{\text{x}}+\frac{2}{2\text{x}-3}=\frac{1}{\text{x}-2},$ $\text{x}\neq0,\frac{3}{2},2$
Answer$\frac{1}{\text{x}}+\frac{2}{2\text{x}-3}=\frac{1}{\text{x}-2},$ $\text{x}\neq0,\frac{3}{2},2$
$\Rightarrow\frac{(2\text{x}-3)+2\text{x}}{\text{x}(2\text{x}-3)}=\frac{1}{(\text{x}-2)}$
$\Rightarrow (x - 2)(4x - 3) = x(2x - 3)$
$\Rightarrow 4x^2 - 11x + 6 = 2x^2 - 3x$
$\Rightarrow 2x^2 - 8x + 6 = 0$
$\Rightarrow x^2 - 4x + 3 = 0$
$\Rightarrow x^2 - 3x - x + 3 = 0$
$\Rightarrow x(x - 3) - (x - 3) = 0$
$\Rightarrow (x - 1)(x - 3) = 0$
$\therefore$ $x - 1 = 0$
$\Rightarrow x = 1$
and $x - 3 = 0$
$\Rightarrow x = 3$
View full question & answer→Question 883 Marks
A passenger train takes one hour less for a journey of $150\ km$ if its speed is increased by 5km/hr from its usual speed. Find the usual speed of the train.
AnswerLet the usual speed of train be x km/hr then
Increased speed of the train = (x +5) km/hr
Time taken by the train under usual speed to cover 150km $=\frac{150}{\text{x}}\text{hr}$
Time taken by the train under increased speed to cove 150km $=\frac{150}{\text{(x+5)}}\text{hr}$
Therefore,
$\frac{150}{\text{x}}-\frac{150}{(\text{x}+5)}=1$
$\frac{\{150(\text{x+5})-150\text{x}\}}{\text{x}(\text{x}+5)}=1$
$\frac{150\text{x}+750-150\text{x}}{\text{x}^2+5\text{x}}=1$
$750 = x^2+ 5x$
$x^2 + 5x - 750 = 0$
$x^2 - 25x + 30x - 750 = 0$
$x(x - 25) + 30(x - 25) = 0$
$(x - 25)(x + 30) = 0$
So, either
$(x - 25) = 0$
$x = 25$
Or$ (x + 30) = 0$
$x = -30 $
But, the speed of the the train can never be negative,
Hence, the usual speed of train is x = 25km/hr
View full question & answer→Question 893 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$2\text{x}^2-2\sqrt{6}\text{x}+3=0$
Answer$2\text{x}^2-2\sqrt{6}\text{x}+3=0$
Here $\text{a}=2,\text{b}=-2\sqrt{6}$ and $\text{c}=3$
$\therefore$ Discriminant $(\text{D})=\text{b}^2-4\text{ac}$
$=(-2\sqrt{6})^2-4\times2\times3$
$=24-24=0$
$\because\text{D}=0$
$\therefore$ Roots are real and equal
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$=\frac{-(-2\sqrt{6})\pm0}{2\times2}$
$=\frac{2\sqrt{6}}{4}=\frac{\sqrt{6}}{2}$ or $\frac{\sqrt{6}}{\sqrt{2}\times\sqrt{2}}=\frac{\sqrt{2}\times\sqrt{3}}{\sqrt{2}\times\sqrt{2}}$
$=\frac{\sqrt{3}}{\sqrt{2}}=\sqrt{\frac{3}{2}}$
$\therefore$ Roots are $=\frac{\sqrt{3}}{\sqrt{2}},\sqrt{\frac{3}{2}}$
View full question & answer→Question 903 Marks
Determine the set of values of k for which the given following quadratic has real root:
$4x^2 - 3kx + 1 = 0$
AnswerThe given quadric equation is $4x^2 - 3kx + 1 = 0$, and roots are real.
Then find the value of k.
Here, $a= 4, b = -3k$ and $c = 1$
As we know that $D = b^2- 4ac$
Putting the value of $a = 4, b = -3k$ and $c = 1$
$= (-3k)^2 - 4 \times 4 \times 1$
$= 9k^2 - 16$
The given equation will have real roots, if $\text{D}\geq0$
$9\text{k}^2-16\geq0$
$9\text{k}^2\geq16$
$\text{k}^2\geq\frac{16}{9}$
$\text{k}\geq\sqrt{\frac{16}{9}}$
$\text{k}\leq-\frac{4}{3}$ or $\text{k}\geq\frac{4}{3}$
Therefore, the value of $\text{k}\leq-\frac{4}{3}$ or $\text{k}\geq\frac{4}{3}$
View full question & answer→Question 913 Marks
A two digit number is $4$ times the sum of its digits and twice the product of its digits. Find the number.
AnswerLet the require digit be $= (10x + y)$
Then according to question,
$(10x + y) = 4(x + y)$
$(10x + y) = 4x + 4y$
$10x + y - 4x - 4y = 0$
$6x - 3y = 0$
$2x - y = 0$
$2x = y ....(i)$
And, $(10x + y) = 2xy ....(ii)$
Now, putting the value of y in equation (ii) from (i)
$(10x + 2x) = 2x \times 2x$
$4x^2 - 12x = 0$
$4x(x - 3) = 0$
$x(x - 3) = 0$
So, either
$x = 0$
Or $(x - 3) = 0$
$x = 3$
So, the digite can never be negative.
When x = 3 then
$y = 2x$
$y = 2 × 3$
$y = 6$
Therefore, number
$= 10x + y$
$= 10 \times 3 + 6$
$= 36$
Thus, the required number be $36$
View full question & answer→Question 923 Marks
Solve the following quadratic equations by factorization:
$\sqrt{2}\text{x}^2-3\text{x}-2\sqrt{2}=0$
AnswerWe have been given
$\sqrt{2}\text{x}^2-3\text{x}-2\sqrt{2}=0$
$\sqrt{2}\text{x}^2-4\text{x}+\text{x}-2\sqrt{2}=0$
$\sqrt{2}\text{x}\big(\text{x}-2\sqrt{2}\big)+1\big(\text{x}-2\sqrt{2}\big)=0$
$\big(\text{x}-2\sqrt{2}\big)\big(\sqrt{2}\text{x}+1\big)=0$
Therefore,
$\text{x}-2\sqrt{2}=0$
$\text{x}=2\sqrt{2}$
or, $\sqrt{2}\text{x}+1=0$
$\sqrt{2}\text{x}=-1$
$\text{x}=\frac{-1}{\sqrt{2}}$
Hence, $\text{x}=2\sqrt{2}$ or $\text{x}=\frac{-1}{\sqrt{2}}$
View full question & answer→Question 933 Marks
The difference of squares of two number is 88. If the larger number is $5$ less than twice the smaller number, then find the two numbers.
AnswerLet teh smaller numbers be x
Then according to question,
The larger number be = 2x - 5, then
$(2x - 5)^2 - x^2 = 88$
$4x^2 - 20x + 25 - x^2 - 88 = 0$
$3x^2 - 20x - 63 = 0$
$3x^2 - 27x + 7x - 63 = 0$
$3x(x - 9) + 7(x - 9) = 0$
$(x - 9)(3x + 7) = 0$
$(x - 9) = 0$
$x = 9$
Or $(3x + 7) = 0$
$\text{x}=\frac{-7}{3}$
Since, x being a positive integer so, x cannot be negative,
Therefore,
When x = 9 then larger number be
$2x - 5 = 2 × 9 - 5$
$2x - 5 = 18 - 5$
$2x - 5 = 13$
Thus, two consecutive number be either $9, 13$
View full question & answer→Question 943 Marks
Find the whole number which when decreased by $20 $is equal to $69$ times the reciprocal of the number.
AnswerLet the whole numbers be x.
Then according to question,
$(\text{x}-20)=69\times\frac{1}{\text{x}}$
$x(x - 20) = 69$
$x^2 - 20x - 69 = 0$
$x^2 - 23x + 3x - 69 = 0$
$x(x - 23) + 3(x - 23) = 0$
$(x - 23)(x + 3) = 0$
$(x - 23) = 0$
$x = 23$
$Or (x + 3) = 0$
$x = -3$
Since, whole numbers being a positive, so x cannot be negative.
Thus, whole numbers be $23$
View full question & answer→Question 953 Marks
Solve the following quadratic equations by factorization:
$3\text{x}^2-2\sqrt{6}\text{x}+2=0$
AnswerWe have been given
$3\text{x}^2-2\sqrt{6}\text{x}+2=0$
$3\text{x}^2-\sqrt{6}\text{x}-\sqrt{6}\text{x}+2=0$
$\sqrt{3}\text{x}\big(\sqrt{3}\text{x}-\sqrt{2}\big)-\sqrt{2}\big(\sqrt{3}\text{x}-\sqrt{2}\big)=0$
$\big(\sqrt{3}\text{x}-\sqrt{2}\big)(\sqrt{3}\text{x}-\sqrt{2})=0$
Therefore,
$\sqrt{3}\text{x}-\sqrt{2}=0$
$\sqrt{3}\text{x}=\sqrt{2}$
$\text{x}=\sqrt{\frac{2}{3}}$
Hence, $\text{x}=\sqrt{\frac{2}{3}}$
View full question & answer→Question 963 Marks
If $p, q$ are real and $p \neq q$, then show that the show that the roots of the equation $(p-q) x^2+5(p+q) x-2(p-q)=0$ are real and unequal.
AnswerThe quadric equation is $(p-q) x^2+5(p+q) x-2(p-q)=0$
Here, $a=(p-q), b=5(p+q)$ and $c=-2(p-q)$
As we know that $D=b^2-4 a c$
Putting the value of $a=(p-q), b=5(p+q)$ and $c=-2(p-q)$
$\Rightarrow D = {5(p + q)}^2 - 4(p - q)(-2(p - q))$
$\Rightarrow D = 25(p^2 + 2pq + q^2) + 8(p^2 - 2pq + q^2)$
$\Rightarrow D = 25p^2+ 50pq + 25q^2 + 8p^2 - 16pq + 8q^2$
$\Rightarrow D = 33p^2 + 34pq + 33q^2$
Since p and q are real and $\text{p}\neq\text{q},$ therefore, the value of $\text{D}\geq0$
Thus, the roots of the given equation are and unequal.
Hence, proved.
View full question & answer→Question 973 Marks
Solve for x.
$\frac{1}{\text{x}-3}-\frac{1}{\text{x}+5}=\frac{1}{6},$ $\text{x}\neq3,-5$
Answer$\frac{1}{\text{x}-3}-\frac{1}{\text{x}+5}=\frac{1}{6},$ $\text{x}\neq3,-5$
$\frac{1}{\text{x}-3}-\frac{1}{\text{x}+5}=\frac{1}{6}$
$\Rightarrow\frac{\text{x}+5-\text{x}+3}{(\text{x}-3)(\text{x}+5)}=\frac{1}{6}$
$\Rightarrow\frac{8}{(\text{x}-3)(\text{x}+5)}=\frac{1}{6}$
$\Rightarrow 48 = x^2 + 2x - 15$
$\Rightarrow x^2 + 2x - 15 - 48 = 0$
$\Rightarrow x^2 + 2x - 63 = 0$
$\Rightarrow x^2 + 9x - 7x - 63 = 0$
$\Rightarrow x(x + 9) - 7(x + 9) = 0$
$\Rightarrow (x - 7)(x + 9) = 0$
$\Rightarrow x = 7, -9$
View full question & answer→Question 983 Marks
The sum of a number and its positive square is $\frac{63}{4},$ find the number.
AnswerLet first numbers be x
Then according to question
$\text{x}+\text{x}^2=\frac{63}{4}$
$4(x + x^2) = 63$
$4x^2 + 4x - 63 = 0$
$4x^2 + 18x - 14x - 63 = 0$
$2x(2x + 9) - 7(2x + 9) = 0$
$(2x + 9)(2x - 7) = 0$
$(2x + 9) = 0$
$\text{x}=-\frac{9}{2}$
Or $(2\text{x}-7)=0$
$\text{x}=\frac{7}{2}$
Thus, the required number be $\frac{7}{2},\frac{-9}{2}$
View full question & answer→Question 993 Marks
The product of two consecutive positive integers is $306$. Form the quadratic equation to find the integers, if x denotes the smaller integer.
AnswerGiven that the smallest integer of 2 consecutive integer is denoted by $x.$
$\Rightarrow $ The two integer will be x and $(x + 1)$
Product of two integers $\Rightarrow x(x + 1)$
Given that the product is 306
$\therefore$ x(x + 1) = 306
$\Rightarrow x^2 + x = 306$
$\Rightarrow x^2 + x - 306 = 0$
$\therefore$ The required quadratic equation is $x^2 + x - 306 = 0$
View full question & answer→Question 1003 Marks
Show that $x = -2$ is a solution of $3x^2 + 13x + 14 = 0.$
AnswerGiven that the equation of $3x^2 + 13x + 14 = 0$
$3x^2 + 7x + 6x + 14 = 0$
$x(3x + 7) + 2(3x + 7) = 0$
$(3x + 7)(x + 2) = 0$
$(3x + 7) = 0$
$\text{x}=\frac{-7}{3}$
Or $(x + 2) = 0$
$x = -2$
Therefore, x = 2 is the solution of given equation.
Hence, proved.
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