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Question: How do you use the quadratic formula to solve for \(x\)-intercepts \({x^2} - 8x + 12 = 0?\)...

How do you use the quadratic formula to solve for xx-intercepts x28x+12=0?{x^2} - 8x + 12 = 0?

Explanation

Solution

In this question using the quadratic formula. First we take the given equation. We identify the value of a,ba, b and cc in the quadratic equation. After that Substitute the values a,ba, b and cc into the quadratic formula and solve for xx. Now simplify the equation, hence we get the equation.
Use the quadratic formula to find the solutions,
b±b24ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
Pull terms out from under the radical, assuming positive real numbers. Now we find the xx value. First we separate the positive and sign values.
Finally we get xx values.

Complete step by step answer:
The given quadratic equation is x28x+12=0{x^2} - 8x + 12 = 0
The quadratic formula states:
For ax2+bx+c=0a{x^2} + bx + c = 0, the value of xx which are the solutions to the equation are given by:
x=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
Where, a=1,b=8a = 1,\,b = - 8 and c=12c = 12
Now, we substitute these values in the quadratic formula
x=(8)±(8)24(1)(12)2(1)x = \dfrac{{ - ( - 8) \pm \sqrt {{{( - 8)}^2} - 4(1)(12)} }}{{2(1)}}
We simplify the equation, hence we get
x=(8)±(8)24(1)(12)2(1)x = \dfrac{{(8) \pm \sqrt {{{( - 8)}^2} - 4(1)(12)} }}{{2(1)}}
Raise8 - 8 to the power of 22
x=(8)±(64)4(1)(12)2(1)x = \dfrac{{(8) \pm \sqrt {(64) - 4(1)(12)} }}{{2(1)}}
Multiply 11by 1212
x=(8)±(64)4×122(1)x = \dfrac{{(8) \pm \sqrt {(64) - 4 \times 12} }}{{2(1)}}
Multiply 44 by 22
x=(8)±64482(1)x = \dfrac{{(8) \pm \sqrt {64 - 48} }}{{2(1)}}
Subtract 4848 from 6464
x=(8)±162(1)x = \dfrac{{(8) \pm \sqrt {16} }}{{2(1)}}
Rewrite 1616 as 42{4^2}
x=(8)±422(1)x = \dfrac{{(8) \pm \sqrt {{4^2}} }}{{2(1)}}
Pull terms out from under the radical, assuming positive real numbers
x=8±42×1x = \dfrac{{8 \pm 4}}{{2 \times 1}}
Multiply 22 by 11
x=8±42x = \dfrac{{8 \pm 4}}{2}
Now we find the xx value. First we separate the positive and negative sign values,
Let,
x=8+42x = \dfrac{{8 + 4}}{2}
Add the numerator, hence we get
x=122x = \dfrac{{12}}{2}
Divide 1212 by 22
x=6x = 6
Let,
x=842x = \dfrac{{8 - 4}}{2}
Subtract the numerator
x=42x = \dfrac{4}{2}
Divide 44 by 22
x=2x = 2
The given equation solution is 66 and 22.
The values of x=6,2x = 6,2

Note: The formula for finding the roots of the quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 is x=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}.
The formula for finding roots of a quadratic equation was known to ancient Babylonians, though not in a form as we derived.
They found the roots by creating the steps as a verse, which is a common practice at their times.
Babylonians used quadratic equations for deciding to choose the dimensions of their land for agriculture.