1 Marks Question

Question 11 Mark

If the quadratic equation $p x^2-2 \sqrt{5} p x+15=0$ has two equal roots, Then find the value of $p$.

Answer

The given quadratic equation is
$p x^2-2 \sqrt{5} p x+15=0$
So, $a=p, b=-2 \sqrt{5} p, c=15$
When the roots of the quadratic equation are equal, then its discriminant will be zero.
$D=b^2-4 a c=0$
$\Rightarrow(-2 \sqrt{5} p)^2-4 \times p \times 15=0$
$\Rightarrow 20 p^2-60 p=0$
$\Rightarrow 20 p(p-3)=0$
$\Rightarrow 20 p=0 \text { or }(p-3)=0$
$\Rightarrow p=0 \text { or } p=3$
$p x^2-2 \sqrt{5} p x+15=0$ is given a quadratic equation,
so, $p$ cannot be equal to $0 .$
Hence, $p=3$

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Question 21 Mark

If $x=-\frac{1}{2}$, is a solution of the quadratic equation $3 x^2+2 k x-3=0$, find the value of $k .$

Answer

It is given that $x=-\frac{1}{2}$ is the solution of the quadratic equation $3 x^2+2 k x-3=0$.
It means it will satisfy the given equation.
Substitute $x=-\frac{1}{2}$ in $3 x^2+2 k x-3=0$ we get,
$3\left(-\frac{1}{2}\right)^2+2 k\left(-\frac{1}{2}\right)-3=0$
$\Rightarrow \frac{3}{4}-k-3=0 $
$\Rightarrow \frac{3-4 k-12}{4}=0$
$\Rightarrow-9-4 k=0 $
$\Rightarrow-4 k=9 $
$\Rightarrow k=-\frac{9}{4}$
Hence, the value of $k$ is $-\frac{9}{4}$.

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Question 31 Mark

Solve the quadratic equation $2 x^2+a x-a^2=0$ for $x$.

Answer

Given $2 x^2+a x-a^2=0$Comparing the given equation with the standard quadratic equation $\left(a x^2+b x+c=0\right)$,
we get $a=2, b=a$ and $c=-a^2$
Using the quadratic formula,
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$, we get:
$x=\frac{-a \pm \sqrt{a^2-4 \times 2\left(-a^2\right)}}{2 \times 2}=\frac{-\alpha \pm \sqrt{9 a^2}}{4}$
$x=\frac{-\alpha+3 a}{4}=\frac{a}{2}$ and $x=\frac{-\alpha-3 a}{4}=-a$
So, the solutions of the given quadratic equation are
$x=\frac{a}{2}$ and $x=-a$

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Question 41 Mark

If $2$ is a zero of the polynomial $a x^2-2 x$, then the value of $'a\ '$ is $......$

Answer

$a x^2-2 x=0$
$x=2$
$a(2)^2-2(2)=0$
$\Rightarrow 4 a-4=0$
$\Rightarrow 4 a =4$
$\Rightarrow a =1$

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Question 51 Mark

Find the value of $k$ for which the quadratic equation $kx ( x -2)+6=0$ has two equal roots.

Answer

$kx(x-2)+6=0$
$\Rightarrow kx^2-2 kx+6=0$
On Comparing $ax ^2+ bx + c =0$
$a=k, b=-2 k, c=6$
$\because$ Roots of the given equation are equal,
$\therefore D=0$
$b^2-4 a c=0$
$\Rightarrow(-2 k)^2-4 \times k \times 6=0$
$\Rightarrow 4 k^2-24 k=0 $
$\Rightarrow 4 k(k-6)=0$
$\Rightarrow k-6=0$
$\Rightarrow k=6$

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