The quadratic equation 3x2 + 45x + 24 = 0 was solved using the Quadratic Formula, x equals negative b plus or minus the square root of b squared minus 4 times a times c, all over 2 times a.

One solution is −14.45. What is the other solution? Round to the hundredths place.

−1.11
−0.55
0.52
14.45

Respuesta :

Answer:

-0.55

Step-by-step explanation:

Quadratic Formula: [tex]\frac{-b\pm\sqrt({b^2}-4ac) }{2a}[/tex]

Case 1: [tex]\frac{-b+\sqrt({b^2}-4ac) }{2a}[/tex]

Substitute values:

[tex]\frac{-45+\sqrt({45^2}-4*3*24) }{2*3}[/tex]

[tex]\frac{-15+\sqrt{193} }{2}[/tex] = -0.55 (2dp) (Answer)

Case 2: [tex]\frac{-b-\sqrt({b^2}-4ac) }{2a}[/tex]

Substitute values:

[tex]\frac{-45-\sqrt({45^2}-4*3*24) }{2*3}[/tex]

[tex]-\frac{15+\sqrt{193}}{2}[/tex] = -14.45 (2dp)

Since x = -14.45 is known, the other solution  is x = -0.55

The two solutions of the given quadratic equation will be -0.55 and -14.45.

What is a quadratic equation?

The polynomial having a degree of two or the maximum power of the variable in a polynomial will be 2 is defined as the quadratic equation and it will cut two intercepts on the graph at the x-axis.

The solution of the given equation will be calculated as:-

3x² + 45x + 24 = 0

a  =  3,  b = 45, and c =24

[tex]= \dfrac{-b^2\pm\sqrt{b^2-4ac}}{2a}[/tex]

[tex]=\dfrac{-45\pm\sqrt{45^2-(4\times 3\times 24)}}{(2\times 3}[/tex]

[tex]=\dfrac{-45\pm41.6}{6}[/tex]

[tex]=\dfrac{-45+41.6}{6}=-0.55[/tex]

[tex]=\dfrac{-45-41.6}{6}=-14.45[/tex]

Therefore two solutions of the given quadratic equation will be -0.55 and -14.45.

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