Factor the trinomial or state that it is prime 4y²+54y-90

Mathematics

2 answers:

lidiya [134]3 years ago

6 0

$4y^2+54y-90=\\
2(2y^2+27y-45)=\\
2(2y^2+30y-3y-45)=\\
2(2y(y+15)-3(y+15))=\\
2(2y-3)(y+15)$

castortr0y [4]3 years ago

4 0

4y²+54y-90
= 2(2y²+27y-45)
= 2(2y-3)(y+15) <- this is the final answer. :)

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Solve The Equation 4x×9y=7 4x-9y=9

hoa [83]

Answer:

$\large\boxed{x=\dfrac{9}{8}-\dfrac{\sqrt{109}}{8},\ y=-\dfrac{1}{2}-\dfrac{\sqrt{109}}{18}}\\or\\\boxed{x=\dfrac{9}{8}+\dfrac{\sqrt{109}}{2},\ y=-\dfrac{1}{2}+\dfrac{\sqrt{109}}{18}}$

Step-by-step explanation:

$\left\{\begin{array}{ccc}4x\times9y=7&(1)\\4x-9y=9&(2)\end{array}\right\\\\(2)\\4x-9y=9\qquad\text{subtract}\ 4x\ \text{from both sides}\\-9y=-4x+9\qquad\text{change the signs}\\9y=4x-9\qquad\text{substitute it to (1)}\\\\4x(4x-9)=7\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\(4x)(4x)+(4x)(-9)=7\\(4x)^2-36x=7\\(4x)^2-2(4x)(4.5)=7\qquad\text{add}\ 4.5^2\ \text{to both sides}\\(4x)^2-2(4x)(4.5)+4.5^2=7+4.5^2\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2$

$(4x-4.5)^2=7+20.25\\(4x-4.5)=27.25\to 4x-4.5=\pm\sqrt{27.25}\\\\4x-\dfrac{45}{10}=\pm\sqrt{\dfrac{2725}{100}}\\\\4x-\dfrac{45}{10}=\pm\dfrac{\sqrt{2725}}{\sqrt{100}}\\\\4x-\dfrac{45}{10}=\pm\dfrac{\sqrt{25\cdot109}}{10}\\\\4x-\dfrac{45}{10}=\pm\dfrac{\sqrt{25}\cdot\sqrt{109}}{10}\\\\4x-\dfrac{45}{10}=\pm\dfrac{5\sqrt{109}}{10}\qquad\text{add}\ \dfrac{45}{10}\ \text{to both sides}\\\\4x=\dfrac{45}{10}\pm\dfrac{5\sqrt{109}}{10}$

$4x=\dfrac{9}{2}\pm\dfrac{\sqrt{109}}{2}\qquad\text{divide both sides by 4}\\\\x=\dfrac{9}{8}\pm\dfrac{\sqrt{109}}{8}\\\\\text{Put the values of}\ x\ \text{to (2):}\\\\9y=4\left(\dfrac{9}{8}\pm\dfrac{\sqrt{109}}{8}\right)-9\\\\9y=\dfrac{9}{2}\pm\dfrac{\sqrt{109}}{2}-\dfrac{18}{2}\\\\9y=-\dfrac{9}{2}\pm\dfrac{\sqrt{109}}{2}\qquad\text{divide both sides by 9}\\\\y=-\dfrac{1}{2}\pm\dfrac{\sqrt{109}}{18}$