Question

A painting with dimensions 10 inches by 14 inches is placed in a picture frame (of constant width), increasing its area to 221 square inches. How many inches is the width of the picture frame?

Answer 1

Answer:

1.5 inches

Step-by-step explanation:

(10+2x)(14+2x) = 221

4x² + 48x - 81 = 0

4x² + 54x - 6x - 81 = 0

2x(2x + 27) - 3(2x + 27) = 0

(2x + 27)(2x - 3) = 0

x = -27/2, 3/2

x = 3/2 or 1½ or 1.5 inches

Answer 2

The width of the picture frame is $\\ x_{2} &=1,5 \end{aligned}$.

Quadratic equation formula

Given:

A painting with dimensions 10 inches by 14 inches is placed in a picture frame (of constant width), increasing its area to 221 square inches.

From the above explanation, we get the equation

$4 x^{2}+2 \cdot 10 x+2 \cdot 14 x+10 \cdot 14 &=221$

$\\ 4 x^{2}+48 x+140 &=221$

By using the quadratic formula, we get

$\\ 4 x^{2}+48 x-81 &=0$

Solve with the quadratic equation formula

For a quadratic equation of the form $a x^{2}+b x+c=0$ the solutions are

$\\ x &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$\\ x &=\frac{-48 \pm \sqrt{48^{2}-4 \cdot 48 \cdot(-81)}}{2 \cdot 4}$

$\\ x &=\frac{-48 \pm \sqrt{3600}}{8}$

$\\ {\left[x_{1}\right.} &=-13,5]$

$\\ x_{2} &=1,5 \end{aligned}$

Therefore, the width of the picture frame is $\\ x_{2} &=1,5 \end{aligned}$.

To learn more about quadratic equation formula

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