Question

11. Find the solutions of x^2-x-30 = 0.

Answer 1

To find the solutions of quadratic equation, there are two ways to do which are:

• Factorize
• Quadratic Formula

Step 1

- Factor the expression.

To factor the expression, refer below:

$\displaystyle \large{ (x - a)(x - b) = {x}^{2} - bx - ax + ab}$

For bx and ax, both can be common-factored. Therefore

$\displaystyle \large{ (x - a)(x - b) = {x}^{2} - (b + a)x + ab}$

From the above, we conclude that:

• The middle term is b+a
• The last term is a×b
• Thus, we have to find two numbers that satisfy a+b and a×b

From the expression, 30 comes from 5×6 and when 5-6 = -1. Therefore, a can be 5 and b can be 6.

$\displaystyle \large{{x}^{2} - x - 30 = (x + 5)(x - 6)}$

Because in the middle term, it is -x which is negative, we have to let the highest number become negative.

From the factored expression:

• The middle term = 5x + (-6x) = -x
• The last term = 5 × (-6) = -30

Then we replace the standard equation with factored form.

$\displaystyle \large{ (x + 5)(x - 6) = 0}$

For this part, we solve like a linear equation where we isolate x. Just think you are solving two linear equations!

Hence

$\displaystyle \large{ x = - 5, 6}$

Therefore, the solutions are x = -5, 6.