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.
$$p2+3p−10 | $$= | $$0 |
$$p2+3p | $$= | $$10 |
$$p2+3p+(32)2 | $$= | $$10+(32)2 |
$$(p+32)2 | $$= | $$10+94 |
$$(p+32)2 | $$= | $$494 |
$$p+32 | $$= | $$±72 |
$$p | $$= | $$±72−32 |
$$p | $$= | $$72−32 and $$−72−32 |
$$p | $$= | $$42 and $$−102 |
$$p | $$= | $$2 and $$−5 |
$$p | $$= | $$2 |
Then | ||
$$x2 | $$= | $$2 |
$$x | $$= | $$±√2 |
AND | ||
$$x2 | $$= | $$−5 |
$$(2x+1)2+2(2x+1)−3 | $$= | $$0 | substitute $$j=2x+1 |
$$j2+2j−3 | $$= | $$0 | |
$$(j+3)(j−1) | $$= | $$0 | |
So | |||
$$j+3 | $$= | $$0 | Where $$j=−3 |
$$j−1 | $$= | $$0 | Where $$j=1 |
Remember that $$j | $$= | $$2x+1 | |
Then | |||
$$j | $$= | $$−3 | becomes |
$$2x+1 | $$= | $$−3 | |
$$2x | $$= | $$−4 | |
$$x | $$= | $$−2 | |
And | |||
$$j | $$= | $$1 | becomes |
$$2x+1 | $$= | $$1 | |
$$2x | $$= | $$0 | |
$$x | $$= | $$0 |
Let's have a look at some other questions.
Solve for $$x: $$x4−20x2+64=0 .
Let $$p be equal to $$x2.
Solve the following equation for $$x:
$$3(9x+10)2+19(9x+10)+20=0
You may let $$p=9x+10.
Consider the equation
$$(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 $$m=()
Solve the equation for $$x by using the substitution $$m=2x.