I can't seem to understand what to do in these following questions, please help.
-3x^2-15x=-42
x^2-2x-8=0
√x+6=x
Thanks for helping out.
What does your book or your teacher ask you to do? Factor them? Solve for x? Or what?
No.2
Solve for x:
x^2-2 x-8=0
The left hand side factors into a product with two terms:
(x-4) (x+2)=0
Split into two equations:
x-4=0 or x+2=0
Add 4 to both sides:
x=4 or x+2=0
Subtract 2 from both sides:
Answer: | x=4 or x=-2
No.3
Solve for x:
6+sqrt(x)=x
Subtract 6 from both sides:
sqrt(x)=x-6
Raise both sides to the power of two:
x=(x-6)^2
Expand out terms of the right hand side:
x=x^2-12 x+36
Subtract x^2-12 x+36 from both sides:
-x^2+13 x-36=0
The left hand side factors into a product with three terms:
-(x-9) (x-4)=0
Multiply both sides by -1:
(x-9) (x-4)=0
Split into two equations:
x-9=0 or x-4=0
Add 9 to both sides:
x=9 or x-4=0
Add 4 to both sides:
x=9 or x=4
6+sqrt(x) ⇒ 6+sqrt(4) = 8
x ⇒ 4 = 4:
So this solution is incorrect
6+sqrt(x) ⇒ 6+sqrt(9) = 9
x ⇒ 9 = 9:
So this solution is correct
The solution is:
Answer: | x=9
+1904 1) \({-3x}^{2}-15x=-42\)
There are many ways to solve this equation. Today, I feel like completing the square so I will do that.
To complete the square, divide each term by the leading coeficent in the first term.
\(\frac{-3x{}^{2}}{-3}-\frac{15x}{-3}=\frac{-42}{-3}\)
\({x}^{2}-(-5x)=14\)
\({x}^{2}+5x=14\)
Now take the term that has the x and divide its coeficent by two and then square it.
\(({\frac{5}{2}})^{2}\)
\(({2.5})^{2}\)
\(6.25\)
Take the answer and add to both side of the equation.
\({x}^{2}+5x+6.25=14+6.25\)
\({x}^{2}+5x+6.25=20.25\)
Factor the left side
\(({x+2.5})^{2}=20.25\)
Take the square root of both sides.
\(\sqrt{({x+2.5})^{2}}=\sqrt{20.25}\)
\(x+2.5=±4.5\)
Seperate the equation into two equations and solve for x.
\(x+2.5-2.5=4.5-2.5\)
\(x+0=2\)
\(x=2\)
\(x+2.5-2.5=-4.5-2.5\)
\(x+0=-7\)
\(x=-7\)
\(x=2\) or \(x=-7\)
Subsitute 2 and -7 for x and check to see if these solutions are true.
\({-3(2)}^{2}-15(2)=-42\)
\(-3(4)-15(2)=-42\)
\(-3(4)-30=-42\)
\(-12-30=-42\)
\(-42=-42\)
\({-3(-7)}^{2}-15(-7)=-42\)
\(-3(49)-15(-7)=-42\)
\(-3(49)-(-105)=-42\)
\(-147-(-105)=-42\)
\(-147+105=-42\)
\(-42=-42\)
Both 2 and -7 work, so those are both solutions.
2) \({x}^{2}-2x-8=0\)
There are many ways to solve this equation. The easiest way to solve this is to factor the left side.
\((x+2)(x-4)=0\)
Solve for x.
\(x+2 = 0\)
\(x+2-2=0-2\)
\(x+0=-2\)
\(x=-2\)
\(x-4=0\)
\(x-4+4=0+4\)
\(x-0=4\)
\(x=4\)
\(x=-2\) or \(x=4\)
Subsitute -2 and 4 for x and check to see if these solutions are true.
\({(-2)}^{2}-2(-2)-8=0\)
\(4-2(-2)-8=0\)
\(4-(-4)-8=0\)
\(4+4-8=0\)
\(8-8=0\)
\(0=0\)
\({4}^{2}-2(4)-8=0\)
\(16-2(4)-8=0\)
\(16-8-8=0\)
\(8-8=0\)
\(0=0\)
Both -2 and 4 work, so those are both solutions.
3) \(\sqrt{x}+6=x\)
Subtract 6 to both sides.
\(\sqrt{x}+6-6=x-6\)
\(\sqrt{x}+0=x-6\)
\(\sqrt{x}=x-6\)
Square both sides.
\({\sqrt{x}}^{2}={(x-6)}^{2}\)
\(x={(x-6)}^{2}\)
Expand the right side
\(x=(x-6)(x-6)\)
\(x={x}^{2}-6x-6x+36\)
\(x={x}^{2}-12x+36\)
Subtract x^2-12x+36 to both sides.
\(x-({x}^{2}-12x+36)={x}^{2}-12x+36-({x}^{2}-12x+36)\)
\(x-{x}^{2}+12x-36={x}^{2}-12x+36-{x}^{2}+12x-36\)
Combine like terms.
\(-{x}^{2}+13x-36={0x}^{2}-0x+0\)
\(-{x}^{2}+13x-36=0-0x+0\)
\(-{x}^{2}+13x-36=0-0+0\)
\(-{x}^{2}+13x-36=0+0\)
\(-{x}^{2}+13x-36=0\)
Divide both sides by -1.
\(\frac{{-x}^{2}}{-1}+\frac{13x}{-1}-\frac{36}{-1}=\frac{0}{-1}\)
\({x}^{2}-13x+36=0\)
You can also multiply both sides by -1 and will get the same result.
\(-{x}^{2}\times-1+13x\times-1-36\times-1=0\times-1\)
\({x}^{2}-13x+36=0\)
Factor the left side.
\((x-9)(x-4)=0\)
Solve for x.
\(x-9=0\)
\(x-9+9=0+9\)
\(x-0=9\)
\(x=9\)
\(x-4=0\)
\(x-4+4=0+4\)
\(x-0=4\)
\(x=4\)
\(x=9\) or \(x=4\)
Subsitute 9 and 4 for x and check to see if these solutions are true.
\(\sqrt{9}+6=9\)
\(3+6=9\)
\(9=9\)
\(\sqrt{4}+6=4\)
\(2+6=4\)
\(8=4\)
Since 9 works and 4 does not work, the only solution is 9.
