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how do i answer 2x-3=sqrroot30-7x
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2x-3=\[\sqrt{30-7x}\]
is this \[2x-3=\sqrt{30-7x}\] OR \[2x-3=\sqrt{30} - 7x\]
the first one
ok soz - I saw your answer after I posted! square both sides of the equation
its okay
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\[(2x-3)^{2}=(\sqrt{30-7x})^{2}\]
so the square root cancels
yes
and 2x-3 turns into 4x^2-12x+9
yes Then collect all terms onto 1 side (other side =0) You then have a quadratic
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okay so ima have 0=4x^2+5x-21
not sure that is correct 4x^2-12x+7x+9-30
so 4x^2-5x-21 ???
yes It doesn't factorise - so use formula
what do i do next??
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use quadratic formula to solve the equation 4x^2-5x-21=0 (ax^2+bx+c=0) x=(-b +/- sqrt (b^2-4ac))/2a
i forgot wAT to do after this part 5x+\[\sqrt{361}\] over 8
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