Question

If -2 is a root of the quadratic equation x2-px-5=0 and the quadratic equation x2+px+k=0 has equal roots find the value of k

Answer 1

Answer:

$\frac{1}{16}$

Step-by-step explanation:

x=(-2)

therefore

$(-2)^{2} -p(-2)-5=0$

$4-5+2p=0$

$p=}\frac{1}{2}$

When the root are equal, then the discriminant(Δ) is 0

Δ = $b^{2} -4ac$

Δ =$(\frac{1}{2}) ^{2} -4k$

Δ = 0

therefore

0 =$\frac{1}{4} - 4k$

$k=\frac{1}{16}$

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Answer 2

Answer:

The value of k for the equal roots of quadratic equation is 0

Step-by-step explanation:

Given as :

The quadratic equation are

x² - p x - 5 = 0

The one of the root of  equation is - 2

According to question

As  - 2 is root of equation, then it must satisfy the given equation

i.e  x² - p x - 5 = 0

we substitute x = - 2 in eq

So, (-2)² - p ( - 2) - 5 = 0

Or, 4 + 2 p - 5 = 0

or, 2 p - 1 = 0

Or, 2 p = 1

∴   p = $\dfrac{1}{2}$

So, The value of p = $\dfrac{1}{2}$

Again

The quadratic equation x² + p x + k = 0  has equal roots

put the value of p in second equation

i.e x² + $\dfrac{x}{2}$  + k = 0

Or, 2 x² + x + 2 k = 0

We know , for equal roots discriminant of quadratic equation = 0

i.e  D = 0

Or, $\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}$  = 0

Or, $\dfrac{-1\pm \sqrt{-1^{2}-4\times 2\times 2k}}{4}$  = 0

Or, ${-1\pm \sqrt{-1^{2}-4\times 2\times 2k}}$ = 0

or, 1 = 1 - 16 k

Or, 16 k = 0

∴  k = 0

Hence , The value of k for the equal roots of quadratic equation is 0 Answer