Question

write all the values of p for which the quadratic equation x2 + px + 16 + = 0 has equal roots. find the root of the equation so obtained​

Answer 1

Step-by-step explanation:

$\huge\bf{\mid{\overline{\underline{ANSWER:-}}\mid}}$

$Let \\ \\ \alpha \: and \: \beta \: be \: roots \: of \: following \: equation \\ \\ Also \: \alpha = \beta \\ \\ {x}^{2} + px + 16 = 0 \\ \\ \alpha + \beta = - \frac{b}{a} = - \frac{p}{1} = - p \\ \\ \alpha \times \beta = \frac{c}{a} = 16 \\ \\ But.. \: \alpha = \beta \\ \\ {\alpha}^{2} = 16 \\ \\ \alpha = \pm 4 = \beta \\ \\ If, \: \alpha = \beta = -4 \: then, \\ \\ \alpha + \beta = p \\ \\ - 4 - 4 = p \\ \\ \boxed{ \implies p = -8 } \\ \\ If, \alpha = \beta = 4 \: then, \\ \\ \alpha + \beta = p \\ \\ 4 + 4 = p \\ \\ \boxed{ \implies p = 8} \\ \\ So, The \: required \: solution \: is \implies \\ \\ \huge\boxed{ p = \pm 8}$

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Answer 2

The possible value of p are -8 and 8

The possible value of roots of given equation are

-4 and -4 or 4 and 4

Step-by-step explanation:

$x^2+px+16=0$

By comparing with quadratic equation

$ax^2+bx+c=0$

We get a=1,b=p,c=16

When roots are equal then

$D=b^2-4ac=0$

Where a=Coefficient of x square

b=Coefficient of x

c=Constant term

Substitute the values then we get

$p^2-4(1)(16)=p^2-64=0$

$p^2-64=0$

$p^2-(8)^2=0$

$(p+8)(p-8)=0$

By using identity $a^2-b^2=(a+b)(a-b)$

$p-8=0$

$p=8$

$p+8=0$

$p=-8$

The possible value of p are -8 or 8

Substitute the value p=-8 then we get

$x^2-8x+16=0$

$x^2-4x-4x+16=0$

$x(x-4)-4(x-4)=0$

$(x-4)(x-4)$

$x-4=0$

$x=4$

Substitute p=8

$x^2+8x+16=0$

$x^2+4x+4x+16=0$

$x(x+4)+4(x+4)=0$

$(x+4)(x+4)=0$

$x+4=0$

$x=-4$

The possible value of roots of given equation are

-4 and -4 or 4 and 4

#Learns more:

https://brainly.in/question/8621278:answered by Tardymanchester