Question

If the root of the equation px^2+qx+r=0 where 2p,q, 2r are in G P, are of the form a^2, 4a-4. Then the value of 2p+4q+7r is

Answer 1

GIVEN :

If the root of the equation $px^2+qx+r=0$ where 2p,q, 2r are in G P, are of the form $a^2$, 4a-4.

TO FIND :

The value of 2p+4q+7r

SOLUTION :

Given 2p,q,2r are in GP their ratios are equal.

⟹ $\frac{q}{2p}=\frac{2r}{q}$

∴ $q^2=4pr$  

Since ,4a−4 are the roots of the equation $px^2+qx+r=0$

As we know that  Difference between the roots of $ax^2+bx+c=0$ is

$\frac{\sqrt{b^2-4ac}}{a}$

⟹$a^2-(4a-4)=\frac{\sqrt{q^2-4pr}}{p}$

$a^2-(4a-4)=\frac{\sqrt{4pr-4pr}}{p}$  (∵ $q^2=4pr$ )

​$a^2-(4a-4)=0$

⟹  $a^2-4a+4=0$

$(a-2)^2=0$

⟹ a=2 is a root of multiplicity 2.

Substitute a=2 in  $a^2$,4a−4 we get,

$2^2$, 4(2)-4

4,8-4

∴ 4,4 are the roots

Sum of the roots=4+4

∴ Sum of the roots=8

Product of the roots=4(4)

∴ Product of the roots=16

For a quadratic equation from the roots the formula is

$x^2-(sum of the roots)x+product of the roots=0$

Substituting the values we get,

Now the equation is $x^2-8x+16=0$

Comparing the above equation with  we get the values of  p=1,q=-8,r=16

Substituting the values in 2p+4q+7r we get,

2p+4q+7r=2(1)+4(-8)+7(16)

=2-32+112

=82

∴ the value of 2p+4q+7r=82