Question

if one root of the quadratic equation 2xsquare + 12x- m= 0is 1.5 times the other root then find the value of m?
1)-27/8
2)-432/25
3)-288/25
4)-16​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

• One root of the quadratic equation 2x² + 12x - m = 0 be 1.5 times the other root.

So, Let assume that

$\rm :\longmapsto\:roots \: are \: \alpha \: and \: 1.5 \: \alpha$

So,

$\rm :\longmapsto\: \alpha , \: 1.5 \alpha \: are \: roots \: of \: {2x}^{2} + 12x - m = 0$

We know that

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm \implies\: \alpha + 1.5 \alpha = - \dfrac{12}{2}$

$\rm \implies\:2.5 \alpha = - 6$

$\rm \implies\:\dfrac{5}{2} \alpha = - 6$

$\bf\implies \: \boxed{ \tt{ \: \alpha = - \: \dfrac{12}{5} \: }}$

Also, We know that

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm \implies\: \alpha \times 1.5 \alpha = \dfrac{ - m}{2}$

$\rm \implies\: \dfrac{3}{2} { \alpha }^{2} = - \dfrac{m}{2}$

$\rm \implies\: {3 \alpha }^{2} = - \: m$

$\rm \implies\:m = - 3 \times {\bigg[\dfrac{ - 12}{5} \bigg]}^{2}$

$\rm \implies\:m = - 3 \times \dfrac{144}{25}$

$\rm \implies\:\boxed{ \tt{ \: m = - \dfrac{432}{25} \: }}$

• So, Option (2) is correct

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More to Know :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac