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# Let $m$ and $n$ be the roots of the quadratic equation $4x^2 + 5x + 3 = 0$. Find $(m + 7)(n + 7)$.

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Let $m$ and $n$ be the roots of the quadratic equation $4x^2 + 5x + 3 = 0$. Find $(m + 7)(n + 7)$.

Guest Feb 22, 2017

#1
+19659
+15

Let m and n be the roots of the quadratic equation 4x^2 + 5x + 3 = 0.

Find (m + 7)(n + 7).

$$\begin{array}{|rcll|} \hline ax^2+bx+c &=& 0 \\ x &=& \frac{ -b \pm \sqrt{b^2-4ac} } {2a} \\\\ x_1 + x_2 &=& \frac{ -b + \sqrt{b^2-4ac} } {2a} + \frac{ -b - \sqrt{b^2-4ac} } {2a} \\ &=& \frac{ -b } {2a}+\frac{\sqrt{b^2-4ac} } {2a} + \frac{ -b } {2a} - \frac{ \sqrt{b^2-4ac} } {2a} \\ &=& \frac{ -2b } {2a} \\ \mathbf{x_1 + x_2} & \mathbf{=} & \mathbf{-\frac{ b } { a}} \qquad \text{ or } \qquad \mathbf{m + n} \mathbf{=} \mathbf{-\frac{ b } { a}} \\\\ x_1\cdot x_2 &=& \left( \frac{ -b + \sqrt{b^2-4ac} } {2a} \right) \cdot \left( \frac{ -b - \sqrt{b^2-4ac} } {2a} \right) \\ &=& \left( \frac{ -b } {2a} + \frac{ \sqrt{b^2-4ac} } {2a} \right) \cdot \left( \frac{ -b } {2a} - \frac{ \sqrt{b^2-4ac} } {2a} \right) \\ &=& \left( \frac{ -b } {2a} \right)^2 - \left( \frac{ \sqrt{b^2-4ac} } {2a} \right)^2 \\ &=& \frac{ b^2 } {4a^2} - \frac{ b^2-4ac } {4a^2} \\ &=& \frac{ b^2-(b^2-4ac) } {4a^2} \\ &=& \frac{ b^2-b^2+4ac } {4a^2} \\ &=& \frac{ 4ac } {4a^2} \\ \mathbf{x_1\cdot x_2} & \mathbf{=} & \mathbf{\frac{ c } { a}} \qquad \text{ or } \qquad \mathbf{m\cdot n} \mathbf{=} \mathbf{\frac{ c } { a}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline && (m + 7)(n + 7) \\ &=& m\cdot n + 7\cdot (m+n) + 7^2 \\ &=& m\cdot n + 7\cdot (m+n) + 49 \\\\ 4x^2 + 5x + 3 &=& 0 \qquad a = 4,\ b=5,\ c=3 \\ m + n &=& -\frac{ b } { a} \\ &=& -\frac{ 5 } { 4 } \\\\ m\cdot n &=& \frac{ c } { a} \\ &=& \frac{ 3 } { 4 } \\\\ (m + 7)(n + 7) &=& m\cdot n + 7\cdot (m+n) + 49 \\ &=& \frac{ 3 } { 4 } + 7\cdot (-\frac{ 5 } { 4 }) + 49 \\ &=& - \frac{ 32 } { 4 } + 49 \\ &=& -8 + 49 \\ &=& 41 \\ \hline \end{array}$$

(m + 7)(n + 7) = 41

heureka  Feb 23, 2017
edited by heureka  Feb 23, 2017
edited by heureka  Feb 23, 2017
#1
+19659
+15

Let m and n be the roots of the quadratic equation 4x^2 + 5x + 3 = 0.

Find (m + 7)(n + 7).

$$\begin{array}{|rcll|} \hline ax^2+bx+c &=& 0 \\ x &=& \frac{ -b \pm \sqrt{b^2-4ac} } {2a} \\\\ x_1 + x_2 &=& \frac{ -b + \sqrt{b^2-4ac} } {2a} + \frac{ -b - \sqrt{b^2-4ac} } {2a} \\ &=& \frac{ -b } {2a}+\frac{\sqrt{b^2-4ac} } {2a} + \frac{ -b } {2a} - \frac{ \sqrt{b^2-4ac} } {2a} \\ &=& \frac{ -2b } {2a} \\ \mathbf{x_1 + x_2} & \mathbf{=} & \mathbf{-\frac{ b } { a}} \qquad \text{ or } \qquad \mathbf{m + n} \mathbf{=} \mathbf{-\frac{ b } { a}} \\\\ x_1\cdot x_2 &=& \left( \frac{ -b + \sqrt{b^2-4ac} } {2a} \right) \cdot \left( \frac{ -b - \sqrt{b^2-4ac} } {2a} \right) \\ &=& \left( \frac{ -b } {2a} + \frac{ \sqrt{b^2-4ac} } {2a} \right) \cdot \left( \frac{ -b } {2a} - \frac{ \sqrt{b^2-4ac} } {2a} \right) \\ &=& \left( \frac{ -b } {2a} \right)^2 - \left( \frac{ \sqrt{b^2-4ac} } {2a} \right)^2 \\ &=& \frac{ b^2 } {4a^2} - \frac{ b^2-4ac } {4a^2} \\ &=& \frac{ b^2-(b^2-4ac) } {4a^2} \\ &=& \frac{ b^2-b^2+4ac } {4a^2} \\ &=& \frac{ 4ac } {4a^2} \\ \mathbf{x_1\cdot x_2} & \mathbf{=} & \mathbf{\frac{ c } { a}} \qquad \text{ or } \qquad \mathbf{m\cdot n} \mathbf{=} \mathbf{\frac{ c } { a}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline && (m + 7)(n + 7) \\ &=& m\cdot n + 7\cdot (m+n) + 7^2 \\ &=& m\cdot n + 7\cdot (m+n) + 49 \\\\ 4x^2 + 5x + 3 &=& 0 \qquad a = 4,\ b=5,\ c=3 \\ m + n &=& -\frac{ b } { a} \\ &=& -\frac{ 5 } { 4 } \\\\ m\cdot n &=& \frac{ c } { a} \\ &=& \frac{ 3 } { 4 } \\\\ (m + 7)(n + 7) &=& m\cdot n + 7\cdot (m+n) + 49 \\ &=& \frac{ 3 } { 4 } + 7\cdot (-\frac{ 5 } { 4 }) + 49 \\ &=& - \frac{ 32 } { 4 } + 49 \\ &=& -8 + 49 \\ &=& 41 \\ \hline \end{array}$$

(m + 7)(n + 7) = 41

heureka  Feb 23, 2017
edited by heureka  Feb 23, 2017
edited by heureka  Feb 23, 2017