Question

question 5 of 10
which of the following are solutions to the equation below?
check all that apply.
$4x^2 + 20x + 25 = 49$

a. $x = \frac{ -\sqrt{7} - 5 }{ 2 }$
b. $x = 1$
c. $x = -6$
d. $x = -1$
e. $x = 6$
f. $x = \frac{ \sqrt{7} - 5 }{ 2 }$

Explanation:

Step1: Rewrite the equation

First, rewrite the given equation \(4x^{2}+20x + 25=49\) as \(4x^{2}+20x+25 - 49=0\), which simplifies to \(4x^{2}+20x - 24 = 0\). We can also notice that the left - hand side of the original equation \(4x^{2}+20x + 25\) is a perfect square trinomial. Recall that \((a + b)^{2}=a^{2}+2ab + b^{2}\), for \(4x^{2}+20x + 25=(2x)^{2}+2\times(2x)\times5+5^{2}=(2x + 5)^{2}\). So the equation becomes \((2x + 5)^{2}=49\).

Step2: Take square roots

Take the square root of both sides of the equation \((2x + 5)^{2}=49\). We know that if \(y^{2}=a\) (\(a\geq0\)), then \(y=\pm\sqrt{a}\). So \(2x+5=\pm\sqrt{49}=\pm7\).

Step3: Solve for x in two cases

Case 1: When \(2x + 5 = 7\)

Subtract 5 from both sides: \(2x=7 - 5=2\). Then divide both sides by 2: \(x = 1\).

Case 2: When \(2x+5=-7\)

Subtract 5 from both sides: \(2x=-7 - 5=-12\). Then divide both sides by 2: \(x=-6\).

We can also solve the quadratic equation \(4x^{2}+20x - 24 = 0\) using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for the quadratic equation \(ax^{2}+bx + c = 0\) (\(a
eq0\)). Here, \(a = 4\), \(b = 20\), \(c=-24\).

First, calculate the discriminant \(\Delta=b^{2}-4ac=(20)^{2}-4\times4\times(-24)=400 + 384 = 784\). Then \(\sqrt{\Delta}=\sqrt{784} = 28\).

\(x=\frac{-20\pm28}{2\times4}=\frac{-20\pm28}{8}\).

For the plus sign: \(x=\frac{-20 + 28}{8}=\frac{8}{8}=1\).

For the minus sign: \(x=\frac{-20-28}{8}=\frac{-48}{8}=-6\).

Now let's check option A and F. If we use the quadratic formula with the original equation \(4x^{2}+20x + 25 - 49=4x^{2}+20x - 24 = 0\), \(a = 4\), \(b = 20\), \(c=-24\). The quadratic formula gives \(x=\frac{-20\pm\sqrt{400+384}}{8}=\frac{-20\pm28}{8}\), not in the form of option A or F. So option A and F are incorrect. Option D (\(x=-1\)): Substitute \(x = - 1\) into the original equation: \(4\times(-1)^{2}+20\times(-1)+25=4-20 + 25=9
eq49\). Option E (\(x = 6\)): Substitute \(x = 6\) into the original equation: \(4\times6^{2}+20\times6+25=4\times36+120 + 25=144+120 + 25=289
eq49\).

Answer:

B. \(x = 1\), C. \(x=-6\)