Question

How do you solve the quadratic using the quadratic formula given \[4{x^2} + 4x - 8 = 1\] ?

Answer 1

Hint: First we need to convert the given equation in the standard quadratic form \[a{x^2} + bx + c = 0\] . A polynomial of degree two is called a quadratic polynomial and its zeros can be found using many methods like factorization, completing the square, graphs, quadratic formula etc. The quadratic formula is used when we fail to find the factors of the equation. In the given question we have to solve the given quadratic equation using the quadratic formula. That is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .

Complete step-by-step answer:
Given \[4{x^2} + 4x - 8 = 1\] .
Now we subtract 1 on both sides we get,
 \[4{x^2} + 4x - 9 = 0\] .
We know the standard quadratic equation is of the form \[a{x^2} + bx + c = 0\] . Comparing this with the given problem we have \[a = 4\] , \[b = 4\] and \[c = - 9\] .
Here we can see that we cannot expand the middle term \[\left( b \right)\] such that whose sum is \[c\] and product is \[a \times c\] .
So we use the quadratic formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .
Substituting the values we have,
 \[ = \dfrac{{ - 4 \pm \sqrt {{4^2} - (4 \times 4 \times - 9)} }}{{2(4)}}\]
 \[ = \dfrac{{ - 4 \pm \sqrt {16 - ( - 144)} }}{8}\]
We know that product of negative and negative will give use positive sign
 \[ = \dfrac{{ - 4 \pm \sqrt {16 + 144} }}{8}\]
 \[ = \dfrac{{ - 4 \pm \sqrt {160} }}{8}\]
But we can write 160 as product of 16 and 10,
 \[ = \dfrac{{ - 4 \pm \sqrt {16 \times 10} }}{8}\]
But 16 is a perfect square and taking it outside we have,
 \[ = \dfrac{{ - 4 \pm 4\sqrt {10} }}{8}\]
Taking 4 as common,
 \[ = \dfrac{{4\left( { - 1 \pm \sqrt {10} } \right)}}{8}\]
 \[ = \dfrac{{ - 1 \pm \sqrt {10} }}{2}\]
Thus we have a two roots,
 \[x = \dfrac{{ - 1 + \sqrt {10} }}{2}\] and \[x = \dfrac{{ - 1 - \sqrt {10} }}{2}\] .
So, the correct answer is “ \[x = \dfrac{{ - 1 + \sqrt {10} }}{2}\] and \[x = \dfrac{{ - 1 - \sqrt {10} }}{2}\] ”.

Note: In various fields of mathematics require the point at which the value of a polynomial is zero, those values are called the factors/solution/zeros of the given polynomial. On the x-axis, the value of y is zero so the roots of an equation are the points on the x-axis that is the roots are simply the x-intercepts. Careful in the calculation part. Since in the above problem we have a positive number inside the root. If we have a negative number inside the root we will get an imaginary number.