Sometimes functions don't even look like quadratics, but with some clever substitutions we can make it look like a quadratic to enable us to solve them.
$p^2+3p-10$p2+3p−10 | $=$= | $0$0 |
$p^2+3p$p2+3p | $=$= | $10$10 |
$p^2+3p+\left(\frac{3}{2}\right)^2$p2+3p+(32)2 | $=$= | $10+\left(\frac{3}{2}\right)^2$10+(32)2 |
$\left(p+\frac{3}{2}\right)^2$(p+32)2 | $=$= | $10+\frac{9}{4}$10+94 |
$\left(p+\frac{3}{2}\right)^2$(p+32)2 | $=$= | $\frac{49}{4}$494 |
$p+\frac{3}{2}$p+32 | $=$= | $\pm\frac{7}{2}$±72 |
$p$p | $=$= | $\pm\frac{7}{2}-\frac{3}{2}$±72−32 |
$p$p | $=$= | $\frac{7}{2}-\frac{3}{2}$72−32 and $\frac{-7}{2}-\frac{3}{2}$−72−32 |
$p$p | $=$= | $\frac{4}{2}$42 and $\frac{-10}{2}$−102 |
$p$p | $=$= | $2$2 and $-5$−5 |
$p$p | $=$= | $2$2 |
Then | ||
$x^2$x2 | $=$= | $2$2 |
$x$x | $=$= | $\pm\sqrt{2}$±√2 |
AND | ||
$x^2$x2 | $=$= | $-5$−5 |
$\left(2x+1\right)^2+2\left(2x+1\right)-3$(2x+1)2+2(2x+1)−3 | $=$= | $0$0 | substitute $j=2x+1$j=2x+1 |
$j^2+2j-3$j2+2j−3 | $=$= | $0$0 | |
$\left(j+3\right)\left(j-1\right)$(j+3)(j−1) | $=$= | $0$0 | |
So | |||
$j+3$j+3 | $=$= | $0$0 | Where $j=-3$j=−3 |
$j-1$j−1 | $=$= | $0$0 | Where $j=1$j=1 |
Remember that $j$j | $=$= | $2x+1$2x+1 | |
Then | |||
$j$j | $=$= | $-3$−3 | becomes |
$2x+1$2x+1 | $=$= | $-3$−3 | |
$2x$2x | $=$= | $-4$−4 | |
$x$x | $=$= | $-2$−2 | |
And | |||
$j$j | $=$= | $1$1 | becomes |
$2x+1$2x+1 | $=$= | $1$1 | |
$2x$2x | $=$= | $0$0 | |
$x$x | $=$= | $0$0 |
Let's have a look at some other questions.
Solve for $x$x: $x^4-20x^2+64=0$x4−20x2+64=0 .
Let $p$p be equal to $x^2$x2.
Solve the following equation for $x$x:
$3\left(9x+10\right)^2+19\left(9x+10\right)+20=0$3(9x+10)2+19(9x+10)+20=0
You may let $p=9x+10$p=9x+10.
Consider the equation
$\left(2^x\right)^2-9\times2^x+8=0$(2x)2−9×2x+8=0
The equation can be reduced to a quadratic equation by using a certain substitution.
By filling in the gaps, determine the correct substitution that would reduce the equation to a quadratic.
Let $m=\left(\editable{}\right)^{\editable{}}$m=()
Solve the equation for $x$x by using the substitution $m=2^x$m=2x.