Question

if both roots are common to the quadratic equation x^2-4=0 and x^2+px-4=0 then p=? a)2 b)0 c)4 d) 1​

Answer 1

Answer:

x=2

Step-by-step explanation:

2^2+p(2)-4=0

4+p2-4=0

2p=0

p=0/2

p=0

Answer 2

Value of p =0.

Option (b) is correct.

Given:

• Two polynomials.
• ${x}^{2} - 4 = 0$ and ${x}^{2} + px - 4 = 0$

To find:

• if both roots are common to the quadratic equation. Then find p.
• a)2
• b)0
• c)4
• d) 1

Solution:

Concept/Formula to be used:

If $\alpha$ and $\beta$ are the zeros of quadratic polynomial

$a {x}^{2} + bx + c = 0$ then

$\alpha + \beta = \frac{ - b}{a}$

Step 1:

Find the zeros of first polynomial.

${x}^{2} - 4 = 0$

or

$(x - 2)(x + 2) = 0$

$\because \: {a}^{2} - {b}^{2} = (a + b)(a - b)$

or

$x - 2 = 0$

or

$\bf x = 2$

or

$x + 2 = 0$

or

$\bf x = - 2$

Zeros of quadratic equation are 2 and -2.

Step 2:

It is given that zeros are common.

Let the zeros of second polynomial be

$\alpha$ and $\beta$

So,

Sum of zeros = -b/a

$\alpha + \beta = \frac{ - p}{1}$

or

$2 - 2 = - p$

or

$- p = 0$

or

$\bf \red{ p = 0}$

Thus,

Value of p =0.

Option (b) is correct.

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