Question

form the quadratic equation if its roots are :- -2 and 11/2​

Answer 1

$\large{\underline{\underline{\red{\sf{\hookrightarrow Solution:- }}}}}$

Given two roots are - 2 and 11/2 are we are required to form a quadratic equⁿ .

So , if -2 and 11/2 will be roots of the quadratic equation [ say p(x) ] , then ,

• x + 2 will be a factor of p(x) .
• x - 11/2 will also be a factor of p(x) .

Now , we can frame the equⁿ like this :

$\large\green{\boxed{\purple{\tt{\pink{\dag} p(x)\:\:=\:\:(x-\alpha)(x-\beta)=0 \pink{\dag}}}}}$

Where alpha and beta are zeroes ,

Now substitute respective values ;

$\tt:\implies p(x)=(x-\alpha)(x-\beta)$

$\tt:\implies (x-\alpha)(x-\beta)=0$

$\tt:\implies (x+2)\bigg(x-\dfrac{11}{2}\bigg)=0$

$\tt:\implies x\bigg(x-\dfrac{11}{2}\bigg)+2\bigg(x-\dfrac{11}{2}\bigg)=0$

$\tt:\implies x^2-\dfrac{11x}{2}+2x-\dfrac{\cancel{22}}{\cancel{2}}=0$

$\tt:\implies 2\bigg( x^2-\dfrac{11x-4x}{2}-11\bigg)=0$

$\underline{\boxed{\purple{\tt{\longmapsto\:\:2x^2-7x-22=0}}}}$

$\pink{\boxed{\purple{\bf{\dag Hence\: quadratic\:equ^n\:is\:(2x^2-7x-22=0)}}}}$

Answer 2

$\huge\sf\pink{Answer}$

☞ The quadratic equation is 2x²-7x-22

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$\huge\sf\blue{Given}$

✭ Roots of an equation are -2 & ¹½

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$\huge\sf\gray{To \:Find}$

◈ The quadratic equation?

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$\huge\sf\purple{Steps}$

$\large\underline{\underline{\sf Concept}}$

So here we shall assume one zero as alpha and another one as beta. So then when we find roots of a equation we finally get it of tue form (x-α)(x-β) where α & β are the roots. Now that we are given the value of α & β we shall substitute them and simply multiply to get our Answer!!

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So we know that,

⪼ $\sf \underline{\boxed{\sf p(x) = (x-\alpha)(x-\beta)}}$

➝ (x-α) = (x-(-2))

➝ (x-α) = (x+2)

Similarly,

➝ (x-β) = (x-¹½)

Substituting these values in the formula,

➳ $\sf p(x) = (x-\alpha)(x-\beta)$

➳ $\sf p(x) = (x+2)\bigg\lgroup x-\dfrac{11}{2}\bigg\rgroup = 0$

➳ $\sf p(x) = x\bigg\lgroup x-\dfrac{11}{2} \bigg\rgroup + x-\dfrac{11}{2} \bigg\rgroup = 0$

➳ $\sf p(x) = \bigg\lgroup x^2-\dfrac{11}{2} x + 2x-11\bigg\rgroup = 0$

So to cancel the fractons we shall multiply the whole equation by 2

➳ $\sf p(x) = 2\bigg\lgroup x^2-\dfrac{11}{2} x + 2x-11\bigg\rgroup$

➳ $\sf p(x) = 2x^2-11x+4x-22$

➳ $\sf \orange{p(x) = 2x^2-7x-22}$

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