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Maths 2 Marks Questions

Quadratic Equations · CBSE STD 10 (CBSE - English Medium). 19 questions.

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19 questions · timed · auto-graded

Question 12 Marks
Find the value of $k$ for which the equation $x^2+k(2 x+k-1)+2=0$ has real and equal roots.
Answer
Simplifying the given equation:
$x^2+k(2 x+k-1)+2=0$
$x^2+2 x k+k^2-k+2=0$
$x^2+2 k x+\left(k^2-k+2\right)=0$
$a=1 ; b=2 k ; c=\left(k^2-k+2\right)$
A quadratic equation has real and equal roots if: $[D=0$ i.e. discriminant is zero$]$
Applying the above condition in given equation:
$D=b^2-4 a c$
$(2 k)^2-4 \times 1 \times\left(k^2-k+2\right)=0$
$4 k^2-4 k^2+4 k-8=0$
$4 k-8=0$
$\Rightarrow 4 k=8$
$\Rightarrow k=\frac{8}{4}=2$
So value of $k=2$
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Question 22 Marks
Find the roots of the quadratic equation $\sqrt{2} x^2+7 x+5 \sqrt{2}=0$
Answer
Factorizing the equation $\sqrt{2} x^2+7 x+5 \sqrt{2}=0$, we'll get
$\Rightarrow \sqrt{2} x^2+2 x+5 x+5 \sqrt{2}=0$
$\Rightarrow \sqrt{2} x(x+\sqrt{2})+5(x+\sqrt{2})=0$
$\Rightarrow(\sqrt{2} x+5)(x+\sqrt{2})=0$
Either $(\sqrt{2} x+5)=0$ or $(x+\sqrt{2})=0$
$\therefore x=-\frac{5}{\sqrt{2}}=-\frac{5 \sqrt{2}}{2} \text { or } x=-\sqrt{2}$
Hence, the roots of the given quadratic equation are $-\frac{5 \sqrt{2}}{2}$ and $-\sqrt{2}$.
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Question 32 Marks
Find the value of $p$ so that the quadratic equation $p x(x-3)+9=0$ has two equal roots.
Answer
We have, $p x(x-3)+9=0$
$\Rightarrow p x^2-3 p x+9=0$
For two equal roots, $D=0$
$\Rightarrow b^2-4 a c=0$
$\Rightarrow(-3 p)^2-4 p(9)=0$
$\Rightarrow 9 p^2-36 p=0$
$\Rightarrow 9 p(p-4)=0$
$\Rightarrow 9 p=0$ and $(p-4)=0$
$\Rightarrow p=0 $ and $p=4$
Here, value of $p$ cannot be $0 .$
Therefore, the value of $p$ is $4 .$
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Question 42 Marks
If $-5$ is a root of the quadratic equation $2 x^2+p x-15=0$ and the quadratic equation $p \left( x ^2+ x \right)+ k =0$ has equal roots, find the value of $k$.
Answer
$-5$ is a root of the quadratic equation $2 x^2+p x-15=0$ so it will satisfy the equation.
$2 \times(-5)^2+p \times(-5)-15=0$
$50-5 p-15=0$
$35-5 p=0$
$p=7$
By putting the value of $p$ in the equation $p\left(x^2+x\right)+k=0$, we get
$7 x^2+7 x+k=0$
This equation has equal roots so its diseriminant will be zero
$D=b^2-4 a c=0$
Here, $a=7, b=7$ and $c=k$
$7^2-4 \times 7 \times k=0$
$49-28 k=0$
$49=28 k$
$k=\frac{49}{28}=\frac{7}{4}$
The value of $k$ is $\frac{7}{4}$
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Question 52 Marks
If $x=\frac{2}{3}$ and $x=-3$ are roots of the quadratic equation $a x^2+7 x+b=0$, find thevalues of $a$ and $b$.
Answer
The roots of the quadratic equation $a x^2+7 x+b=0$ are $-3$ and $\frac{2}{3}$.
The sum of roots $=\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^2}$
$\Rightarrow-3+\frac{2}{3}=\frac{-(7)}{a}$
$\Rightarrow \frac{-9+2}{3}=\frac{-7}{a}$
$\Rightarrow \frac{-7}{3}=\frac{-7}{a}$
$\Rightarrow a=3 \ldots \text { (i) }$
Also, we know that the product of roots
$=\frac{\text { Constant term }}{\text { Coefficient of } x^2}$
$\Rightarrow-3 \times \frac{2}{3}=\frac{b}{a}$
$\Rightarrow-2=\frac{b}{3}($ Using $(i))$
$\Rightarrow b=-6$
Hence, the values of $a$ and $b$ are $3$ and $-6$ respectively.
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Question 62 Marks
Solve the following quadratic equations for $x :4 x^2+4 b x-\left(a^2-b^2\right)=0$
Answer
The given quadratic equation is
$4 x^2+4 b x-\left(a^2-b^2\right)=0$
So, $a=4, b=4 b, c=-\left(a^2-b^2\right)$
Quadratic formula to find the roots is
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$x=\frac{-4 b \pm \sqrt{(4 b)^2-4 \times 4 \times\left\{-\left(a^2-b^2\right)\right\}}}{2 \times 4}$
$x=\frac{-4 b \pm \sqrt{16 b^2+16\left(a^2-b^2\right)}}{2 \times 4}$
$x=\frac{-4 b \pm \sqrt{16 b^2+16 a^2-16 b^2}}{8}$
$x=\frac{-4 b \pm \sqrt{16 a^2}}{8}$
$x=\frac{-4 b \pm 4 a}{8}$
$x=\frac{-b \pm a}{2}$
$x=\frac{-b+a}{2}, \frac{-b-a}{2}$
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Question 72 Marks
Solve the following quadratic equation for $x : 4 x^2-4 a^2 x+\left(a^4-b^4\right)=0$
Answer
$4 x^2-4 a^2 x+\left(a^4-b^4\right)=0$
This equation is of the form
$a x^2+b x+c=0$
Here $a=4, b=-4 a^2, c=\left(a^4-b^4\right)$
The quadratic formula to solve for $x$ is
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$\Rightarrow x=\frac{4 a^2 \pm \sqrt{\left(-4 a^2\right)^2-4(4)\left(a^4-b^4\right)}}{2(4)}$
$\Rightarrow x=\frac{4 a^2 \pm \sqrt{16 a^4-16 a^4+16 b^4}}{8}$
$\Rightarrow x=\frac{4 a^2 \pm \sqrt{16 b^4}}{8} \Rightarrow x=\frac{4 a^2 \pm 4 b^2}{8}$
Dividing the numerator and denominator by $4$ we get,
$x=\frac{a^2 \pm b^2}{2}$
$\Rightarrow x=\frac{a^2+b^2}{2} \text { or } x=\frac{a^2-b^2}{2}$
Hence, the values of $x$ are $\frac{a^2+b^2}{2}$ and $\frac{a^2-b^2}{2}$.
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Question 82 Marks
Find the values of $p$ for which the quadratic equation $4 x^2+p x+3=0$ has equal roots
Answer
Consider the equation:
$4 x^2+p x+3=0$
For roots to be equal, discriminant is equal to zero.
$b^2-4 a c=0$
Here $a=4, b=p, c=3$
Substitute the values in the equation
$\Rightarrow p^2-4(4)(3)=0$
$\Rightarrow p^2-48=0$
$\Rightarrow p^2=48$
$\Rightarrow p=\sqrt{48}$
$=4 \sqrt{3}$
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Question 92 Marks
Find the quadratic polynomial whose zeroes are $\sqrt{3}+\sqrt{5}$ and $\sqrt{5}-\sqrt{3}$.
Answer
Quadratic polynomial in terms of $x$, where coefficient of $x$ is sum of zeros and constant term will be product of zeros.
$\Rightarrow x^2-(\sqrt{3}+\sqrt{5}+\sqrt{5}-\sqrt{3}) x+(\sqrt{3}+\sqrt{5})(\sqrt{5}-\sqrt{3})$
$\Rightarrow x^2-(2 \sqrt{5}) x+(5-3) $
$\Rightarrow x^2-2 \sqrt{5} x+2$
Hence the required quadratic polynomial is
$\Rightarrow x^2-2 \sqrt{5} x+2$
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Question 102 Marks
Solve the following quadratic equation for: $4 \sqrt{3} x^2+5 x-2 \sqrt{3}=0$
Answer
$4 \sqrt{3} x^2+5 x-2 \sqrt{3}=0$
$4 \sqrt{3} x^2+8 x-3 x-2 \sqrt{3}=0$
Factoring out the common terms,
$4 x(\sqrt{3} x+2)-\sqrt{3}(\sqrt{3} x+2)=0$
$(4 x-\sqrt{3})(\sqrt{3} x+2)=0$
Equating both the factors to
$4 x-\sqrt{3}=0$
$\Rightarrow x=\frac{\sqrt{3}}{4}$
And $\sqrt{3} x+2=0$
$
\Rightarrow x=\frac{-2}{\sqrt{3}}
$
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Question 112 Marks
Form a quadratic polynomial whose zeroes are $\frac{3-\sqrt{3}}{5}$ and $\frac{3+\sqrt{3}}{5}$
Answer
$\frac{3-\sqrt{3}}{5}$ and $\frac{3+\sqrt{3}}{5}$ are the zeroes of the quadratic polynomial.
Let $\alpha=\frac{3-\sqrt{3}}{5}$ and $\beta=\frac{3+\sqrt{3}}{5}$
Sum of zeroes $=\alpha+\beta$
$\Rightarrow \frac{3-\sqrt{3}}{5}+\frac{3+\sqrt{3}}{5}=\frac{3-\sqrt{3}+3+\sqrt{3}}{5}=\frac{6}{5}$
Product of zeroes $=\alpha \beta$
$\Rightarrow\left(\frac{3-\sqrt{3}}{5}\right)\left(\frac{3+\sqrt{3}}{5}\right)=\frac{9-3}{25}=\frac{6}{25}$
The quadratic equation whose zeroes are $\alpha$ and $\beta$ are:
$\Rightarrow x^2-(\alpha+\beta) x+\alpha \beta=0$
$\Rightarrow x^2-\left(\frac{6}{5}\right) x+\left(\frac{6}{25}\right)=0$
Multiply both sides by
$25 x^2-30 x+6=0$
Therefore, the required polynomial is
$25 x^2-30 x+6=0$
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Question 122 Marks
Find the value of $p$ for which the roots of the equation $p x(x-2)+6=0$, are equal.
Answer
The given equation is
$p x(x-2)+6=0$
$\Rightarrow p x^2-2 p x+6=0$
Here, $a=p, b=-2 p$ and $c=6$.
$\therefore D=b^2-4 a c$
$=(-2 p)^2-4 \times p \times 6$
$=4 p^2-24 p$
The given equation will have equal roots, if
$D=0$
$\Rightarrow 4 p^2-24 p=0$
$\Rightarrow 4 p(p-6)=0$
$\Rightarrow p=0 \text { or } p=6$
Here, $p=0$ is not possible because putting $p=0$ in the equation will eliminate the term containing $x^2$.
Thus, the value of $p$ is $6 .$
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Question 132 Marks
If the sum of two natural numbers is $8$ and their product is $15 ,$ find the numbers.
Answer
Let one natural number bex.
Therefore, another natural number will be $8-x$.
Now, It is given that the product of these two natural numbers is $15 .$
Thus, $\Rightarrow 8 x-x^2=15$
$\Rightarrow x^2-8 x+15=0$
$\Rightarrow x^2-3 x-5 x+15=0$
$\Rightarrow x(x-3)-5(x-3)=0$
$\Rightarrow(x-3)(x-5)=0$
$\Rightarrow x-3=0 \text { or } x-5=0$
$\Rightarrow x=3 \text { or } x=5$
Therefore, if first natural number is $3 ,$ the second number will be $5.$
And, if first natural number is $5 ,$ the second number will be $3 .$
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Question 142 Marks
Solve the following quadratic equations for $x: x^2-4 a x-b^2+4 a^2=0$
Answer
We have, $x^2-4 a x-b^2+4 a^2=0$
$\Rightarrow\left(x^2-4 a x+4 a^2\right)-b^2=0$
$\Rightarrow(x-2 a)^2-b^2=0$
$\Rightarrow(x-2 a+b)=0 \text { or }(x-2 a-b)=0$
$\Rightarrow(x-2 a+b)(x-2 a-b)=0$
$\Rightarrow x=2 a-b \text { or } x=2 a+b$
Thus, $2 a-b$ and $2 a+b$ are the two roots of the equation $x^2-4 a x-b^2+4 a^2=0$.
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Question 152 Marks
Solve the quadratic equation $x^2+2 \sqrt{2 x}-6=0$ for $x$.
Answer
Given quadratic equation is:
$x^2+2 \sqrt{2 x}-6=0$
$\Rightarrow x^2+3 \sqrt{2 x}-\sqrt{2 x}-6=0$
$\Rightarrow x(x+3 \sqrt{2})-\sqrt{2}(x-3 \sqrt{2})=0$
$\Rightarrow(x+3 \sqrt{2})(x-\sqrt{2})=0$
$\Rightarrow x+3 \sqrt{2}=0 \text { or } x-\sqrt{2}=0$
$\Rightarrow x=-3 \sqrt{2} \text { or } x=\sqrt{2}$
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Question 162 Marks
Find the value of $' k\ '$ so that the quadratic equation $3 x^2-5 x-2 k=0$ has real and equal roots.
Answer
Given quadratic equation is
$3 x^2-5 x-2 k=0$
For real and equal roots $D =0$;
$\text { i.e., } b^2-4 a c=0$ on comparing the given equation with
$ax^2+bx+c=0,$  wet get 
$a=3, b=-5 $ and  $c=-2 k$
$\therefore b^2-4 ac=0$
$\Rightarrow(-5) 2-4(3)(-2 k)=0$
$\Rightarrow 25+24 k=0$
$\Rightarrow k=-\frac{25}{24}$
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Question 172 Marks
Find the roots of the quadratic equation $x ^2- x -2=0$.
Answer
Given,
$
\Rightarrow \quad x^2-x-2=0
$
The given equation is in the form of, $a x^2+b x+c=0$
$
\begin{array}{ll}
So, \Rightarrow & a=1 \\
\Rightarrow & b=-1 \\
\Rightarrow & c=-2
\end{array}
$
Finding the discriminant of the equation,
$
\Rightarrow \quad D=b^2-4 a c
$
Putting the values of $a, b$ and $c$,
$
\begin{array}{ll}
\Rightarrow & D=(-1)^2-4(1)(-2) \\
\Rightarrow & D=9 \\
\Rightarrow & D>0
\end{array}
$
Since, D is positive, so, the equation has two distinct real roots.
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Question 182 Marks
Find the discriminant of the quadratic equation $3 x^2-2 x+\frac{1}{3}=0$ and hence find the nature of its roots.
Answer
$3 x^2-2 x+\frac{1}{3}=0$
$\frac{3 \times 3 x^2-3 \times 2 x+1}{3}=0$
$9 x^2-6 x+1=0 \times 3$
$9 x^2-6 x+1=0$
Comparing equation with
$a x^2+b x+c=0$
$a=9, b=-6, c=1$
We know that
$D=b^2-4 ac$
$D=(-6)^2-4 \times 9 \times 1$
$D=36-36$
$D=0$
Since $D =0$
The given equation has two equal real roots
Now using quadratic formula to find roots
$x=\frac{-b \pm \sqrt{D}}{2 a}$
$x=\frac{-(-6) \pm \sqrt{0}}{2 \times 9}$
$x=\frac{6+0}{18}$
$x=\frac{6}{18}$
$x=\frac{1}{3}$
Hence, the roots of the equation are $\frac{1}{3}, \frac{1}{3}$.
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Question 192 Marks
Find the roots of the quadratic equation $3 x^2-4 \sqrt{3} x+4=0$
Answer
$3 x^2-4 \sqrt{3} x+4=0$
$\Rightarrow 3 x^2-2 \sqrt{3} x-2 \sqrt{3} x+4=0$
$\Rightarrow 3 x(\sqrt{3} x-2)-2(\sqrt{3} x-2)=0$
$\Rightarrow(\sqrt{3} x-2)(\sqrt{3} x-2)=0$
$\Rightarrow x=\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}$
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