Question

Use the quadratic formula to solve x² + 9x + 10 = 0.

What are the solutions to the equation?

Round irrational solutions to the nearest tenth.



A) x=−8.9 and x=−0.14

B) x=−7.7 and x=−1.3

C) x=−7 and x=−1

D) x=−10 and x = 2

Answer 1

a = 1 b = 9 c = 10
x =  [-b +- sq root(b^2 -4*a*c)] / 2*a
x = [-9 +- sq root (81 -40) ] / 2
x = -9 /2 + sq root (41) / 2
x = -4.5 + 3.202
= -1.298

x = -4.5  - 3.202
= -7.702

answer is B

Answer 2

The solutions to the equation are -7.7 and -1.3 option (B) is correct.

What is a quadratic equation?

Any equation of the form $\rm ax^2+bx+c=0$  where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

We have a quadratic equation:

$\rm x^2+9x+10=0$

Comparing with the standard form of a quadratic equation, we get:

a = 1, b = 9, and c = 10

We know the formula for finding the roots of a quadratic equation:

$\rm x = \frac{-b\pm\sqrt{b^2-4ac} }{2a}$

Putting the values of the a, b, and c in the formula, we get:

$\rm x = \frac{-9\pm\sqrt{9^2-4(1)(10)} }{2\times1}$

$\rm x = \frac{-9\pm\sqrt{81-40} }{2}\\\\\rm x = \frac{-9\pm\sqrt{41} }{2}$

Taking the plus sign first:

$\rm x = \frac{-9+\sqrt{41} }{2} \Rightarrow -1.298 \approx -1.3$

Now taking the negative sign:

$\rm x = \frac{-9-\sqrt{41} }{2} \Rightarrow -7.7015 \approx -7.7$

Thus, the solutions to the equation are -1.3 and -7.7 option (B) is correct.

Learn more about quadratic equations here:

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