Question

Please solve each equation for x. 1.) x² -14x² +49=0 2.) x² +8x² +16=0 3.) x² -32=0

Answer 1

Step-by-step explanation:

1 .) x^2 -7x -7x +49=0

x(x-7)-7(x-7)=0

(x-7) (x-7)=0

x =7

2.) x^2+4x+4x+16=0

(x+4)(x+4)=0

x =. -4

3) x^2 -32=0

x^2 =32

x =4√2

Answer 2

$\star\:\:\bf\large\underline\blue{Questions:-}$

Solve the equations.

$\star\:\:\bf\large\underline\blue{Solutions:-}$

$\star\:\:\bf\underline\blue{Question\:1:-}$

${x}^{2} - 14x + 49 = 0$

$\implies {x}^{2} - 7x - 7x + 49 = 0$

$\implies x(x - 7) - 7(x - 7) = 0$

$\implies (x - 7)(x - 7) = 0$

Therefore, either $x-7=0$ or $x-7=0$ i.e., roots are equal.

So,

$x - 7 = 0$

$\implies x = 7$

Also,

$x - 7 = 0$

$\implies x = 7$

Therefore, Roots are:-

• 7 and
• 7

$\star\:\:\bf\underline\blue{Question\:2:-}$

${x}^{2} + 8x + 16 = 0$

$\implies {x}^{2} + 2 \times x \times 4 + {4}^{2} = 0$

$\implies (x + 4)(x + 4) = 0$Therefore, either $x+4=0$ or $x+4=0$

Hence,

$x + 4 = 0$

$\implies x = 4$

Also,

$x + 4 = 0$

$\implies x = 4$

Therefore, Roots are:-

• 4 and
• 4

$\star\:\:\bf\underline\blue{Question\:3:-}$

${x}^{2} - 32 = 0$

$\implies {x} = \sqrt{32}$

$\implies {x} = \pm 4\sqrt{2}$

Hence,

$x = 4 \sqrt{2}$

And,

$x = - 4 \sqrt{2}$

Therefore, Roots are:-

• $4 \sqrt{2}$
• $- 4 \sqrt{2}$