factorizing quadratic equation
2x2 +13x +15=0
(2x +___)(x+___) = 0
\(\begin{array}{|lrcll|} \hline (2x + a)(x+ b) &=& 0 \\ 2x^2 + 2xb+ax+ab &=& 0 \\ 2x^2 + x\underbrace{(a+2b)}_{=13} + \underbrace{ab}_{=15} &=& 0 \quad |& \quad \text{compare with } 2x^2 +13x +15=0 \\\\ (1) & ab &=& 15 \\ & b &=& \frac{15}{a} \\\\ (2) & a+2b &=& 13 \\ & a + \frac{2\cdot 15}{a} \\ & a + \frac{30}{a} &=& 13 \quad | \quad \cdot a \\ & a^2 + 30 &=& 13a \\ & a^2 -13a + 30 &=& 0 \\\\ & (a-10)(a-3) &=& 0 \\\\ & \mathbf{a_1 = 10} & \text{and} & \mathbf{a_2 = 3} \\\\ & b_1 = \frac{15}{10 } && b_2 = \frac{15}{3} \\ & \mathbf{b_1 = 1.5} && \mathbf{ b_2 = 5} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline (2x + 10)(x+1.5) = 0 \\ \text{ or } \\ (2x + 3)(x+ 5 ) = 0 \\ \hline \end{array}\)
2x2 +13x +15=0 (2x+ )(x+ )
\(2x^2 +13x +15=0\)
a b c
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(x = {-13 \pm \sqrt{169-4\cdot 2\cdot 15} \over 2\cdot 2}\)
\(x=\frac{-13\pm\sqrt{49}}{4}\)
\(x_1=-\frac{3}{2}\\ x_2=-5\)
\((x+5)(x+\frac{3}{2})=x^2+\frac{3}{2}x+5x+\frac{15}{2}\\ (x+5)(x+\frac{3}{2})=x^2+\frac{13}{2}x+\frac{15}{2}\)
\(2(x+5)(x+\frac{3}{2})=2x^2 +13x +15\)
\((x+5)(2x+3)=2x^2 +13x +15\)
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