Question

Solve the equation \[{{x}^{2}}+5x-\left( {{a}^{2}}+a-6 \right)=0\].

Answer 1

Hint: Roots of any quadratic $A{{x}^{2}}+Bx+C=0$ are given by quadratic formula, given as
$x=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}$

Compare the given quadratic with the general quadratic, mentioned above i.e. $A{{x}^{2}}+Bx+C$. And hence, get A, B, C and put it in the equation expressed above for getting roots.

Use: - ${{\left( A+B \right)}^{2}}={{A}^{2}}+{{B}^{2}}+2AB$

Complete step-by-step answer:

Given expression in the problem is

\[{{x}^{2}}+5x-\left( {{a}^{2}}+a-6 \right)=0\] ……………………………………….(i)

As we can observe the given expression in equation (i) is a quadratic in ‘x’.

So, as we can get roots of any quadratic $A{{x}^{2}}+Bx+C=0$, using the quadratic formula given as,

$x=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}$ …………………………………………….(ii)

So, we can get value of ‘x’ of equation (i) with the help of equation (ii) as,

$x=\dfrac{-5\pm \sqrt{{{\left( 5 \right)}^{2}}-4\times 1\times \left( -\left( {{a}^{2}}+a-6 \right) \right)}}{2\times 1}$

$x=\dfrac{-5\pm \sqrt{25+4\left( {{a}^{2}}+a-6 \right)}}{2}$

$x=\dfrac{-5\pm \sqrt{25+4{{a}^{2}}+4a-24}}{2}$

$x=\dfrac{-5\pm \sqrt{4{{a}^{2}}+4a+1}}{2}$ …....................................(iii)

Now, we know the algebraic identity of ${{\left( A+B \right)}^{2}}$ is given as

${{\left( A+B \right)}^{2}}={{A}^{2}}+{{B}^{2}}+2AB$ ……………………………………(iv)

Now, the expression given in square root can be simplified with respect to (iv) as

$4{{a}^{2}}+4a+1={{\left( 2a \right)}^{2}}+2\times 2a\times 1+{{\left( 1 \right)}^{2}}$
Om comparing this expression with equation (i), we get

 $A=2a$, $B=1$

Hence, we get

$2\times 2a\times 1+{{1}^{2}}={{\left( 2a+1 \right)}^{2}}$

$\Rightarrow 4{{a}^{2}}+4a+1={{\left( 2a+1 \right)}^{2}}$ ……………………..(v)

Now, we can replace $4{{a}^{2}}+4a+1$ by $2a+1$ in the equation (iii). We get

$x=\dfrac{-5\pm \sqrt{{{(2a+1)}^{2}}}}{2}$

As, square root of any number is equal to that number. So, we get value of ‘x’ as

$x=\dfrac{-5\pm \left( 2a+1 \right)}{2}$

Now, we can get two values of x as

$x=\dfrac{-5+\left( 2a+1 \right)}{2}$ or $x=\dfrac{-5-\left( 2a+1 \right)}{2}$

$x=\dfrac{-5+2a+1}{2}$ or $x=\dfrac{-5-2a-1}{2}$

$x=\dfrac{2a-4}{2}$ or $x=\dfrac{-2a-6}{2}$

$x=a-2$ or $x=-a-3$

Hence, roots of the given equation are $a-2$ and $-a-3$.


Note: One may try to solve the given quadratic with mid-term splitting method as well. So, he/she first do the factorization of ${{a}^{2}}+a-6$ in following way: -

${{x}^{2}}+5x-\left( {{a}^{2}}+3a-2a-6 \right)=0$

${{x}^{2}}+5x-\left( a\left( a+3 \right)-2\left( a+3 \right) \right)=0$

${{x}^{2}}+5x-\left( a-2 \right)\left( a+3 \right)=0$

Now, split 5 as $-\left( a-2 \right)$ and $\left( a+3 \right)$.

Use the quadratic formula very carefully. Don’t change the position of any value, otherwise the answer will be wrong. Calculation is also an important part of this question, so take care of it as well.