Question

please solve this question​

Answer 1

⇒ The given equation is ∣x∣

2

+∣x∣−6=0 ---- ( 1 )

⇒ When x≥0, then ∣x∣=x, so equation ( 1 ) becomes,

x

2

+x−6=0

⇒ x

2

+3x−2x−6=0

⇒ x(x+3)−2(x+3)=0

⇒ (x+3)(x−2)=0

⇒ x=−3,2

When x<0, then ∣x∣=−x, so equation ( 1 ) becomes,

x

2

−x−6=0

⇒ x

2

−3x+2x−6=0

⇒ x(x−3)+2(x−3)=0

⇒ (x−3)(x+2)=0

⇒ x=3,−2

⇒ Sum of the roots =−3+2+3−2=0

∴ Here, roots are real and equal.

Answer 2

$\large \sf \underbrace{ \underline{Solution+Explanation}}$

First we have to find roots of equation

➠x²-5x+6

➠x²-2x-3x+6

➠x(x-2)-3(x-2)

➠(x-3)(x-2)

So roots are 3 and 2

which means a=3 and b=2

$\sf We \: have \: to \: find \: equation \: of \: roots \: \frac{a}{2} and \: \frac{b}{2}$

$\sf i.e. \frac{3}{2} \: and \: \frac{2}{2}$

As we know that formula to find quadratic equation using roots is

x²-(sum)x+product

$\sf ➠\:\:x²-(\frac{3}{2}+1)x+(\frac{3}{2}×1)=0$

$\sf➠\:\: x²-(\frac{3+2}{2})x+\frac{3}{2}=0$

$\sf ➠\:\:x²-(\frac{5}{2})x+\frac{3}{2}=0$

$\sf multiply\: this\: equation \:by\: 2$

➠ 2x²-5x+3=0

So this is the required equation.