Solve. 16x^{2}-25=0 x=\Box

Solve.
$16{x}^{2}-25=0$
$x=B\otimes$
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Jenny Sheppard
Step 1
Given Data:
Equation: $16{x}^{2}-25=0$
First, apply factorization in the given equation then solve for the values of x.
The useful factorization is,
$\left({a}^{2}-{b}^{2}\right)=\left(a-b\right)\left(a+b\right)$
Step 2
Solve the given equation,
$16{x}^{2}-25=0$
${\left(4x\right)}^{2}-{\left(5\right)}^{2}=0$
(4x-5)(4x+5)=0
$x=\frac{5}{4}$ & $x=-\frac{5}{4}$
Thus, the values of x are $-\frac{5}{4}$ and $\frac{5}{4}$ (or -1.25 and 1.25).

puhnut1m
We will first write the equation in standard form $a{x}^{2}+bx+c=0$ to identify a,b and c.
Where a is the coefficient of the ${x}^{2}$ term, b is the coefficient of the x term and c is the constant term.
Given equation: $16{x}^{2}-25=0$
The variable in the equation is x: $x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
Use a=16, b=0 and c=-25: $x=\frac{-\left(0\right)±\sqrt{{\left(0\right)}^{2}-4\cdot 16\cdot \left(-25\right)}}{2\left(16\right)}$
$x=\frac{-0±\sqrt{0+1600}}{32}$
$x=\frac{±\sqrt{0+1600}}{32}$
Use $\sqrt{1600}=40:x=\frac{±40}{32}$
We know $\frac{a±b}{c}$ means $\frac{a+b}{c}$ or $\frac{a-b}{c}:x=\frac{42}{30}$ or $x=-\frac{40}{32}$
Simplify: $x=\frac{5}{4}$ or $x=-\frac{5}{4}$
Hence the roots are $-\frac{5}{4}$ and $\frac{5}{4}$