Question

solve the following quadratic equations. Leave surds in your answers.

a) (x-3)^2-3=0

b) (3x-7)^2 =8

2a) x^4-5x^2+4=0

b) x-8=2squareroot x

Answer 1

a) $(x-3)^{2}-3=0$

Add $3$ to both sides:

$(x-3)^{2}-3+\textcolor{primary}{3}=0+\textcolor{primary}{3}$

Add like terms:

$(x-3)^{2}=3$

Take the square root for both sides:

$x-3=\pm \sqrt{3}$

Add $3$ to both sides:

$x=3\pm \sqrt{3}$

The solutions are $x=3- \sqrt{3}$ and $x=3+\sqrt{3}$

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b)$(3x-7)^{2}=8$

Take the square root for both sides:

$3x-7=\pm \sqrt{8}$

Add $7$ to both sides:

$3x=7\pm \sqrt{8} ​$

Divide both sides by $3$:

$x=\frac{7\pm\sqrt{8}}{3}$

Simplify the expression $\sqrt{8}$ as $2\sqrt{2}$ :

$x=\frac{7\pm2\sqrt{2}}{3}$

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2a)$x^4-5x^2+4=0$

Find two numbers whose product is $4$ "the third term" and have the sum of $-5$ "coefficient of the second term":

The numbers are $-4$ and $-1$, such that $-4 \times -1 =4$ and $-4+(-1)=-5$

Now factor the expression:

$(x^2-1)(x^2-4)=0$

Using Zero-Product Property, it follows that:

$(x^2-1)=0$ or $(x^2-4)=0$

Take the first equation:

$x^2-1=0$

Add $1$ to both sides:

$x^2=1$

Take the square root for both sides:

$x=\pm1$

Take the second equation:

$x^2-4=0$

Add $4$ to both sides:

$x^2=4$

Take the square root for both sides:

$x=\pm2$

Therefore, the solutions are $-1,1,-2, \text{and}~ 2$

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b)$x-8=2\sqrt{x}$

Square both sides:

$(x-8)^2=(2\sqrt{x})^2$

Expand the square of the binomial on the left-hand side:

$x^2-16x+64=(2\sqrt{x})^2$

Square the right-hand side:

$x^2-16x+64=4x$

Subtract $4x$ from both sides:

$x^2-16x+64-4x=4x-4x$

Add like terms:

$x^2-20x+64=0$

Write $-20x$ as a difference:

$x^2-4x-16x+64=0$

Factor out $x$ from the first two expressions:

$x(x-4)-16x+64=0$

Factor out $-16$ from the second two expressions:

$x(x-4)-16($ $x-4)$ $=0$

Factor out $(x-4)$ from the expression:

$(x-4)(x-16)=0$

Using Zero-Product Property, it follows that:

$(x-4)=0~~\text{or}~~(x-16)=0$

$x=4~~\text{or}~~x=16$