Key to "Howe do you solve x2−12x+52=0?"

High School
Algebra
Algebra I
Quadratic function and equation

Isis Solis

Answered question

2022-01-20

Howe do you solve

x^{2} - 12 x + 52 = 0

Answer & Explanation

Karlie Wilkins

Beginner2022-01-21Added 7 answers

Step 1
$x^{2} - 12 x + 52 = 0$
The equation is of the form $a x^{2} + b x + c = 0$ where:
$a = 1, b = - 12, c = 52$
The Discriminant is given by:
$Δ = b^{2} - 4 \times a \times c$
$= {(- 12)}^{2} - (4 \times 1 \times 52)$
$= 144 - 208$
$= - 64$
If $Δ = 0$ then there is only one solution.
(for $Δ > 0$ there are two solutions, for $Δ < 0$ there are no real solutions)
As $Δ = - 64$ , this equation has no real solutions
Note:
The solutions are normally found using the formula
$x = \frac{- b \pm \sqrt{Δ}}{2 \times a}$
As $Δ = - 64, x = \frac{- (- 12) \pm \sqrt{- 64}}{2 \times 1} = \frac{12 \pm \sqrt{- 64}}{2}$

Rosa Nicholson

Beginner2022-01-22Added 13 answers

Step 1
$x^{2} - 12 x + 52$
is of the form $a x^{2} + b x + c$ with $a = 1, b = - 12$ and $c = 52$ . This has discriminant given by the formula:
$Δ = b^{2} - 4 a c = {(- 12)}^{2} - (4 \times 1 \times 52)$
$= 144 - 208 = - 64 < 0$
So $x^{2} - 12 x + 52 =$ has no real roots. It has two distinct comples roots.

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