Question

if the roots of a quadratic equation x²-px+4=0 are equal then p=?​

Answer 1

Answer:

a=1,b=-p,c=4

b2-4ac=0

(-p)2-4(1)(4)=0

p2-16=0

p2=16

p=4

Answer 2

$\large{\underline{\underline{\mathfrak{\green{\bf{QUESTION:-}}}}}}$.

• if the roots of a quadratic equation x²-px+4=0 are equal then p=?

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$\lage{\underline{\underline{\mathfrak{\bf{Given\:Here:-}}}}}$.

• equation $\:(x^2-px+4)\:=\:0$

• Roots are equal

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$\large{\underline{\underline{\mathfrak{\bf{Find\:Here:-}}}}}$.

• Value of p = ?

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$\large{\underline{\underline{\mathfrak{\</u><u>b</u><u>f{</u><u>E</u><u>X</u><u>P</u><u>L</u><u>A</u><u>N</u><u>A</u><u>T</u><u>I</u><u>O</u><u>N</u><u>:-}}}}}}$.

Let ,

• $\alpha\:and\:\beta$ are Roots of this equations

So,

We know that's

$\large\boxed{\boxed{\:Sum\:of\:Roots\:=\frac{-(Coefficient\:of\:x)}{(Cofficient\:of\:x^2)}}}$

$\implies\:(\alpha\:+\beta)\:=\frac{-(-p)}{1}$

$\implies\:(\alpha\:+\beta)\:=\:p$

But, A/C to question ,

• Roots are equal

So,Take here

• $\alpha\:=\:\beta$

$\implies\:(\alpha\:+\alpha)\:=\:p$

$\implies\:(2\alpha)\:=\:p$

$\large\implies\boxed{\boxed{\:(\alpha)\:=\frac{p}{2}}}$....(1)

Again , we know that's

$\large\boxed{\boxed{\:Product\:of\:roots\:=\frac{(Constant\:part)}{(Coefficient\:of\:x^2)}}}$

$\implies\:(\alpha\beta)\:=\frac{4}{1}$

$\implies\:(\alpha\beta)\:=\:4$

But, we know ,

• $\alpha\:=\beta$

So,

$\implies\:(\alpha\alpha)\:=\:4$

$\implies\:(\alpha)^2\:=\:4$

$\implies\:(\alpha)\:=\:±\:\sqrt{4}$

$\implies\boxed{\boxed{\:(\alpha)\:=\:±2}}$

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Case (1):-

• When $\alpha\:=\:2$

Keep value of $\alpha$ in (1) ,

$\implies\:2\:=\frac{p}{2}$

$\implies\:p\:=\frac{2}{2}$

$\implies\boxed{\boxed{\:p\:=\:1}}$

Case (2):-

• When $\alpha\:=\:-2$

Keep value in (1) ,

$\implies\:(-2)\:=\frac{p}{2}$

$\implies\:p\:=\frac{-2}{2}$

$\implies\boxed{\boxed{\:p\:=\:-1}}$

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$\large{\underline{\underline{\mathfrak{\bf{Answer:-}}}}}$.

$\boxed{\:Value\:Of\:p\:= \:-1\:and\:1}$

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