Question

If the roots of the given quadratic equation are real and equal then find the value of ‘m’.

(m-12) x² + 2 (m-12) x + 2 ​

Answer 1

ANSWER:

Given:

• p(x) = (m - 12)x² + 2(m - 12)x + 2
• Roots of p(x) are real and equal

To Find:

• Value of m

Solution:

$:\longrightarrow p(x)= (m-12)x^2+2(m-12)x+2\\\\\text{Here, coefficient of x ^2 =(m-12), coefficient of x=2(m-12) & constant=2}\\\\\text{So, on comparing it with general form: ax ^2 +bx+c }\\\\:\implies\text{a=(m-12), \:\:\: b=2(m-12)=2m-24, \:\:\: c=2}\\\\\text{We know that the Quadratic Formula:}\\\\:\implies x=\dfrac{-b\pm\sqrt{D}}{2a}\:\:(D=b^2-4ac)\\\\\text{Also, the roots of p(x) are real and equal so,}\\\\:\implies D=0\\\\:\implies(b^2-4ac)=0\\\\\text{Placing values of a, b and c,}\\\\:\implies(2m-24)^2-4(m-12)(2)=0$

$\text{\underline{METHOD 1}}\\\\:\implies(2m-24)^2-4(m-12)(2)=0\\\\:\implies(2m-24)^2=4(2m-24)\\\\\text{Cancelling, one (2m-24) on LHS,}\\\\:\implies2m-24=4\\\\:\implies2m=28\\\\\bf{:\implies m=14}$

$\text{\underline{METHOD 2}}\\\\:\implies(2m-24)^2-4(m-12)(2)=0\\\\\text{We know that, (a-b)^2=a^2-2ab+b^2 . So,}\\\\:\implies(4m^2-96m+576)-8m+96=0\\\\:\implies4m^2-96m+576-8m+96=0\\\\:\implies4m^2-104m+672=0\\\\:\implies4(m^2-26m+168)=0\\\\:\implies m^2-12m-14m+168=0\\\\:\implies m(m-12)-14(m-12)=0\\\\:\implies(m-12)(m-14)=0\\\\:\implies m=14\:\:or\:\:12.\\\\\text{But, putting m=12, will make (m-12)=0. This in turn will make a=0.}\\\\\text{And if a=0 then, it is not a quadratic polynomial. So,}\\\\\bf{:\implies m=14}$

Formulae Used:

• Quadratic Formula: x=(-b±√D)/2a. Here, D=b^2-4ac
• (a-b)^2 = a^2 + b^2 - 2ab