Question

Find the value of k for which the roots of the quadratic
equation 2x²+Kx+8=0, will have equal value​

Answer 1

$\LARGE{\bf{\underline{\underline{GIVEN:-}}}}$

• The rots of the equation 2x²+kx+8=0 are equal.

$\LARGE{\bf{\underline{\underline{TO:FIND:-}}}}$

• The value of k.

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

Since the roots are equal, the discriminant will be equal to zero. In other words:

$\mapsto \sf{\red{b^2 - 4ac=0}}$

In the equation 2x²+kx+8=0, It is in the form of ax²+bx+c where:

• b = k.
• a = 2.
• c = 8.

Substitute the values:

$\sf \to k^2- 4 \times 2 \times 8=0$

$\sf \to k^2-64=0$

$\sf \to k^2=64$

$\sf \to k=\sqrt{64}$

$\sf \to k=\pm8$

∴K is either +8 or -8

Answer 2

Question

find the value of K for which the root of the quadratic equation 2x^2 + kx + 8 =0, will have equal value .

Answer

Given,

$= > 2 {x}^{2} + kx + 8 = 0$

$\bold{ \underline{ \: to \: find \: }}$

the value of k in the given equation .

• we have to first determine in the equation by " ax^2+ bx+ C =0 " that which is

1. a
2. b
3. C

Now,

according to the equation

A = 2

B = k

C = 8

Now,

$= {k}^{2} - 4 \times 2 \times 8 = 0$

$= {k}^{2} - 64 = 0$

$= > {k}^{2} = 64 \\ = > k = \sqrt{64} \\ = > k = 8$

So, the value of k will be +8 or -8 .