Question

Solve the following quadraticequation (x^2 + 5x +4) (x^2 +5x + 6) = 120​

Answer 1

ANSWER:

To Solve:

• (x² + 5x + 4)(x² + 5x + 6) = 120

Solution:

We are given that,

$\implies \left(x^2+5x+4\right)\left(x^2+5x+6\right)=120$

Now, let us assume that x² + 5x = t.

So,

$\implies \left(x^2+5x+4\right)\left(x^2+5x+6\right)=120$

$\implies (t+4)(t+6)=120$

$\implies t(t+6)+4(t+6)=120$

$\implies t^2+6t+4t+24=120$

$\implies t^2+10t+24-120=0$

$\implies t^2+10t-96=0$

$\implies t^2+16t-6t-96=0$

$\implies t(t+16)-6(t+16)=0$

$\implies (t+16)(t-6)=0$

So,

$\implies t=-16, 6$

We had, x² + 5x = t.

Case 1) Taking t = -16

$\implies x^2+5x=-16$

$\implies x^2+5x+16=0$

Using Quadratic eqn,

$\implies x=\dfrac{(-5)\pm\sqrt{(5)^2-4(1)(16)}}{2(1)}$

$\implies x=\dfrac{-5\pm\sqrt{25-64}}{2}$

$\implies x=\dfrac{-5\pm\sqrt{-39}}{2}$

As, D < 0, No real roots exists.

Case 2) Taking t = 6

$\implies x^2+5x=6$

$\implies x^2+5x-6=0$

Using Quadratic eqn,

$\implies x=\dfrac{(-5)\pm\sqrt{(5)^2-4(1)(-6)}}{2(1)}$

$\implies x=\dfrac{-5\pm\sqrt{25+24}}{2}$

$\implies x=\dfrac{-5\pm\sqrt{49}}{2}$

$\implies x=\dfrac{-5\pm7}{2}$

$\implies x=\dfrac{-5+7}{2},\:\:x=\dfrac{-5-7}{2}$

$\implies x=\dfrac{2}{2},\:\:x=\dfrac{-12}{2}$

Hence,

$\implies\bf x=1,\: -6$

Therefore, value of x is 1 & -6.