Question

How do you find the zeros, real and imaginary of a $y=-7{{x}^{2}}-15x-35$ using the quadratic formula?

Answer 1

Hint: We will write the given equation in the standard form. We will use the quadratic formula given by \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] to find the real and imaginary zeros of a quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c.$

Complete step by step solution:
Let us consider the given problem. We are asked to find the real and imaginary zeros of the given quadratic equation $y=-7{{x}^{2}}-15x-35$ using the quadratic formula.
The given equation is in the standard form.
Suppose that we are given with a quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c.$ The solution of this quadratic equation can be found by using the formula given by \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.\] This formula is called the quadratic formula. The equation is the general form of a standard quadratic equation.
If we compare the given equation with the general equation, then we will get \[a=-7,b=-15\] and $c=-35.$
Now, let us use the quadratic formula by applying these values in it for finding the solution of the given quadratic equation.
So, the quadratic formula for finding the solution of the given quadratic equation is given by, \[x=\dfrac{-\left( -15 \right)\pm \sqrt{{{\left( -15 \right)}^{2}}-4\left( -7 \right)\left( -35 \right)}}{2\left( -7 \right)}.\]
Let us do the calculation to find the value of $x.$
In the first step, we will get \[x=\dfrac{15\pm \sqrt{225-980}}{-14}.\]
Now, we will get \[x=\dfrac{15\pm \sqrt{-755}}{-14}.\]
We know that$-755=-1\times 755$ and $\sqrt{-1}=i.$
Therefore, we will get \[x=\dfrac{15\pm i\sqrt{755}}{-14}.\]
Therefore, the solutions of the given quadratic equation are \[x=\dfrac{15+i\sqrt{775}}{-14}\] or \[x=\dfrac{15-i\sqrt{775}}{-14}.\]

Hence the given quadratic equation has two complex solutions and no real solutions.

Note: We can use the determinant test to find if an equation has real distinct roots, real double roots and complex roots. We will consider the terms inside the square root. That is, \[{{b}^{2}}-4ac.\] If \[{{b}^{2}}-4ac<0,\] then the equation has two distinct real roots. If \[{{b}^{2}}-4ac=0,\] then the equation has double roots. if \[{{b}^{2}}-4ac>0,\] then the equation has two complex roots.