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Wilma and Greg were trying to solve the quadratic equation
\[x^2 + bx + c = 0.\]Wilma wrote down the wrong value of $b$ (but her value of $c$ was correct), and found the roots to be $1$ and $6.$ Greg wrote down the wrong value of $c$ (but his value of $b$ was correct), and found the roots to be $-1$ and $-4.$ What are the actual roots of $x^2 + bx + c = 0$?

Respuesta :

Answer:

The roots are -2 and -3

Step-by-step explanation:

In the function x² + bx + c, a is equal to 1, then the factored form of this function is: (x - x1)*(x - x2) where x1 and x2 are the roots.

Wilma got the following expression:

(x - 1)*(x - 6) = x² - 6x - x + 6 = x² - 7x + 6, where '+6' is correct but '-7x' is incorrect.

Greg got the following expression:

(x + 1)*(x + 4) = x² + 4x + x + 4 = x² + 5x + 4, where '+5x' is correct but '+4' is incorrect.

Therefore the quadratic equation is: x² + 5x + 6. Its roots are:

[tex]x = \frac{-b \pm \sqrt{b^2 - 4(a)(c)}}{2(a)} [/tex]

[tex]x = \frac{-5 \pm \sqrt{5^2 - 4(1)(6)}}{2(1)} [/tex]

[tex]x = \frac{-5 \pm 1}{2}[/tex]

[tex]x_1 = \frac{-5 + 1}{2}[/tex]

[tex]x_1 = -2 [/tex]

[tex]x_2 = \frac{-5 - 1}{2}[/tex]

[tex]x_2 = -3 [/tex]

batolisis

The actual roots of [tex]x^2+bx+c=0[/tex] are [tex]-2[/tex] and [tex]-3[/tex]

Wilma wrongly reported the roots as [tex]x=1 \text{ or }x=6[/tex]

[tex]\implies(x-1)(x-6)=0\\\implies x^2-7x+6=0[/tex]

Greg wrongly reported the roots as [tex]x=-1 \text{ or }x=-4[/tex]

[tex]\implies(x+1)(x+4)=0\\\implies x^2+5x+4=0[/tex]

To get the actual roots, since Greg's value of b [tex](+5)[/tex] was correct, and Wilma's value of c [tex](+6)[/tex] was correct, we then have the correct quadratic equation as

[tex]x^2+5x+6=0\\\implies (x+2)(x+3)=0\\\implies x=-2 \text{ or } x=-3[/tex]

Therefore the actual roots are  [tex]-2[/tex] and [tex]-3[/tex]

Learn more about roots of quadratic equation here: https://brainly.com/question/21435490

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