Question

Find the values of k for which the following equations have real roots

(i) (ii)

(iii) $x^2-4kx+k=0$ (iv) $kx\left(x-2\sqrt{5}\right)+10=0$

(v) $kx(x-3)+9=0$ (vi) $4x^2+kx+3=0$

Answer 1

(i) The given quadric equation is , and roots are real and equal

Then find the value of k.

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if D = 0

Therefore, the value of

(ii) The given quadric equation is , and roots are real and equal

Then find the value of k.

Here,

So,

As we know that

Putting the value of

The given equation will have real and equal roots, if D = 0

Now factorizing of the above equation

So, either

Therefore, the value of

(iii) The given quadratic equation is $x^2-4kx+k=0$, and roots are real and equal.

Then find the value of k.

Here,

$x^2-4kx+k=0$

So,

$a=1,b=-4k\mathrm{and}c=k.$

As we know that $D=b^2-4ac$

Putting the value of $a=1,b=-4k\mathrm{and}c=k.$

$$\begin{gathered} D={\left(-4k\right)}^2-4\left(1\right)\left(k\right) \\ =16k^2-4k \end{gathered}$$

The given equation will have real and equal roots, if D = 0.

So, $16k^2-4k=0$

Now factorizing the above equation,

$$\begin{gathered} 16k^2-4k=0 \\ \Rightarrow 4k\left(4k-1\right)=0 \\ \Rightarrow 4k=0\mathrm{or}4k-1=0 \\ \Rightarrow k=0\mathrm{or}k=\frac{1}{4} \end{gathered}$$

Therefore, the value of $k=0,\frac{1}{4}$.

(iv) The given quadratic equation is $kx\left(x-2\sqrt{5}\right)+10=0$, and roots are real and equal.

Then find the value of k.

Here,

$$\begin{gathered} kx\left(x-2\sqrt{5}\right)+10=0 \\ \Rightarrow k{x}^2-2\sqrt{5}kx+10=0 \end{gathered}$$

So,

$a=k,b=-2\sqrt{5}k\mathrm{and}c=10.$

As we know that $D=b^2-4ac$

Putting the value of $a=k,b=-2\sqrt{5}k\mathrm{and}c=10.$

$$\begin{gathered} D={\left(-2\sqrt{5}k\right)}^2-4\left(k\right)\left(10\right) \\ =20k^2-40k \end{gathered}$$

The given equation will have real and equal roots, if D = 0.

So, $20k^2-40k=0$

Now factorizing the above equation,

$$\begin{gathered} 20k^2-40k=0 \\ \Rightarrow 20k\left(k-2\right)=0 \\ \Rightarrow 20k=0\mathrm{or}k-2=0 \\ \Rightarrow k=0\mathrm{or}k=2 \end{gathered}$$

Therefore, the value of $k=0,2$.

(v) The given quadratic equation is $px(x-3)+9=0$, and roots are real and equal.

Then find the value of p.

Here,

$$\begin{gathered} px(x-3)+9=0 \\ \Rightarrow p{x}^2-3px+9=0 \end{gathered}$$

So,

$a=p,b=-3p\mathrm{and}c=9.$

As we know that $D=b^2-4ac$

Putting the value of $a=p,b=-3p\mathrm{and}c=9.$

$$\begin{gathered} D={\left(-3p\right)}^2-4\left(p\right)\left(9\right) \\ =9p^2-36p \end{gathered}$$

The given equation will have real and equal roots, if D = 0.

So, $9p^2-36p=0$

Now factorizing the above equation,

$$\begin{gathered} 9p^2-36p=0 \\ \Rightarrow 9p\left(p-4\right)=0 \\ \Rightarrow 9p=0\mathrm{or}p-4=0 \\ \Rightarrow p=0\mathrm{or}p=4 \end{gathered}$$

Therefore, the value of $p=0,4.$

(vi) The given quadratic equation is $4x^2+px+3=0$, and roots are real and equal.

Then find the value of p.

Here,

$4x^2+px+3=0$

So,

$a=4,b=p\mathrm{and}c=3.$

As we know that $D=b^2-4ac$

Putting the value of $a=4,b=p\mathrm{and}c=3.$

$$\begin{gathered} D={\left(p\right)}^2-4\left(4\right)\left(3\right) \\ =p^2-48 \end{gathered}$$

The given equation will have real and equal roots, if D = 0.

So, $p^2-48=0$

Now factorizing the above equation,

$$\begin{gathered} p^2-48=0 \\ \Rightarrow p^2-{\left(4\sqrt{3}\right)}^2=0 \\ \Rightarrow \left(p-4\sqrt{3}\right)\left(p+4\sqrt{3}\right)=0 \\ \Rightarrow p-4\sqrt{3}=0\mathrm{or}p+4\sqrt{3}=0 \\ \Rightarrow p=4\sqrt{3}\mathrm{or}p=-4\sqrt{3} \end{gathered}$$

Therefore, the value of $p=\pm 4\sqrt{3}.$